QPaperGen

Explanation: As \(\text{18}{{\text{0}}^{\text{o}}}\text{= }\!\!\pi\!\!\text{ }\) radian Therefore \({{1}^{\circ }}=\dfrac{\pi }{180}\) radian hence, \(\text{2}{{\text{5}}^{\text{o}}}\text{=}\dfrac{\text{ }\!\!\pi\!\!\text{ }}{\text{180}}\text{ }\!\!\times\!\!\text{ 25}\) radian \(\text{=}\dfrac{\text{5 }\!\!\pi\!\!\text{ }}{\text{36}}\)radian

Explanation: \(\frac{121π}{
540}\)

Explanation: 343°38′ 11″ (approximately)

Explanation: 35.7 cm

Explanation: 6.28 cm

Solution not available.

Solution not available.

Solution not available.

Solution not available.

Solution not available.

Solution not available.

Solution not available.

Solution not available.

Solution not available.

Solution not available.

Solution not available.

Explanation:
\[ \text{Given:}\quad
\cos x = -\frac{1}{2}, \quad x \text{ lies in the third quadrant}
\]

Using the Pythagorean identity:
\[
\sin^2 x = 1 - \cos^2 x = 1 - \left(-\frac{1}{2}\right)^2 = \frac{3}{4}
\]
\[
\sin x = -\frac{\sqrt{3}}{2}
\]
(in third quadrant, sine is negative)
\[\]

Other functions:
\[
\tan x = \frac{\sin x}{\cos x} = \sqrt{3}
\]
\[
\csc x = \frac{1}{\sin x} = -\frac{2}{\sqrt{3}}
\]
\[
\sec x = \frac{1}{\cos x} = -2
\]
\[
\cot x = \frac{\cos x}{\sin x} = \frac{1}{\sqrt{3}}
\]

Explanation: Given:\(
\sin x = \frac{3}{5}, \quad x \text{ lies in the second quadrant}
\)\[\]

Since \(\sin x = \frac{3}{5}\):

\[
\sin^2 x + \cos^2 x = 1
\]
\[
\left(\frac{3}{5}\right)^2 + \cos^2 x = 1
\]
\[
\frac{9}{25} + \cos^2 x = 1
\]
\[
\cos^2 x = 1 - \frac{9}{25} = \frac{16}{25}
\]
\[
\cos x = -\frac{4}{5}
\]
(Negative in the second quadrant)

\[
\tan x = \frac{\sin x}{\cos x} = \frac{3/5}{-4/5} = -\frac{3}{4}
\]

\[
\cot x = \frac{\cos x}{\sin x} = \frac{-4/5}{3/5} = -\frac{4}{3}
\]

\[
\sec x = \frac{1}{\cos x} = -\frac{5}{4}
\]

\[
\csc x = \frac{1}{\sin x} = \frac{5}{3}
\]

Summary Table:

\[
\begin{array}{|c|c|}
\hline
\text{Function} & \text{Value} \\
\hline
\sin x & \dfrac{3}{5} \\
\cos x & -\dfrac{4}{5} \\
\tan x & -\dfrac{3}{4} \\
\cot x & -\dfrac{4}{3} \\
\sec x & -\dfrac{5}{4} \\
\csc x & \dfrac{5}{3} \\
\hline
\end{array}
\]

Explanation: Given:
\[
\sec x = \frac{13}{5}, \quad x \text{ lies in the fourth quadrant}
\]

Since \(\sec x = \frac{1}{\cos x}\):
\[
\cos x = \frac{5}{13}
\]
In the fourth quadrant, cosine is positive, sine is negative.

\(\textbf{Find \(\sin x\):}\) by using
\[
\sin^2 x + \cos^2 x = 1
\]
\[
\sin^2 x = 1 - \cos^2 x = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}
\]
\[
\sin x = -\frac{12}{13}
\]
(Negative in fourth quadrant)
\[\]

Other trigonometric functions:
\[
\tan x = \frac{\sin x}{\cos x} = \frac{-12/13}{5/13} = -\frac{12}{5}
\]
\[
\csc x = \frac{1}{\sin x} = -\frac{13}{12}
\]
\[
\cot x = \frac{\cos x}{\sin x} = \frac{5/13}{-12/13} = -\frac{5}{12}
\]

Summary Table:
\[
\begin{array}{|c|c|}
\hline
\text{Function} & \text{Value} \\
\hline
\sin x & -\frac{12}{13} \\
\cos x & \frac{5}{13} \\
\tan x & -\frac{12}{5} \\
\sec x & \frac{13}{5} \\
\csc x & -\frac{13}{12} \\
\cot x & -\frac{5}{12} \\
\hline
\end{array}
\]

Explanation:
We know that
\[
\sin\frac{\pi}{6} = \frac{1}{2}, \quad \cos\frac{\pi}{3} = \frac{1}{2}, \quad \tan\frac{\pi}{4} = 1.
\]

Substituting these values:
\[
\sin^2\frac{\pi}{6} + \cos^2\frac{\pi}{3} - \tan^2\frac{\pi}{4}
= \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^2 - (1)^2
\]

\[
= \frac{1}{4} + \frac{1}{4} - 1
= \frac{1}{2} - 1 = -\frac{1}{2}.
\]

Explanation: We know the standard values:
\[
\sin\frac{\pi}{6} = \frac{1}{2}, \quad \cos\frac{\pi}{3} = \frac{1}{2}.
\]

Also,
\[
\text{cosec}\left(\frac{7\pi}{6}\right) = \text{cosec}\left(\pi + \frac{\pi}{6}\right) = \text{cosec}\left(\frac{\pi}{6}\right) = -2.
\]

Therefore,
\[
\text{LHS} = 2\sin^2\frac{\pi}{6} + \text{cosec}^2\frac{7\pi}{6}\,\cos^2\frac{\pi}{3}
\]

\[
= 2\left(\frac{1}{2}\right)^2 + \left((-2)^2\right)\left(\frac{1}{2}\right)^2
\]

\[
= 2\cdot \frac{1}{4} + 4\cdot \frac{1}{4}
\]

\[
= \frac{1}{2} + 1 = \frac{3}{2}.
\]

\[
\therefore \text{LHS} = \text{RHS}.
\]

Hence proved.

Explanation: We know the standard values:
\[
\cot\frac{\pi}{6} = \sqrt{3}, \quad \tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}.
\]

Also,
\[
\text{cosec}\left(\frac{5\pi}{6}\right) = \text{cosec}\left(\pi - \frac{\pi}{6}\right) = \text{cosec}\left(\frac{\pi}{6}\right) = 2.
\]

Now substituting these values in the LHS,
\[
\text{LHS} = \cot^2\frac{\pi}{6} + \text{cosec}\frac{5\pi}{6} + 3\tan^2\frac{\pi}{6}
\]

\[
= (\sqrt{3})^2 + 2 + 3\left(\frac{1}{\sqrt{3}}\right)^2
\]

\[
= 3 + 2 + 3\cdot \frac{1}{3}
\]

\[
= 3 + 2 + 1 = 6.
\]

\[
\therefore \text{LHS} = \text{RHS}.
\]

Explanation:

We evaluate each term:

\begin{align*}
\sin\left(\frac{3\pi}{4}\right) &= \sin\left(\pi - \frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \\
\Rightarrow \sin^2\left(\frac{3\pi}{4}\right) &= \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \\
\\
\cos\left(\frac{\pi}{4}\right) &= \frac{1}{\sqrt{2}} \\
\Rightarrow \cos^2\left(\frac{\pi}{4}\right) &= \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \\
\\
\sec\left(\frac{\pi}{3}\right) &= \frac{1}{\cos\left(\frac{\pi}{3}\right)} = \frac{1}{\frac{1}{2}} = 2 \\
\Rightarrow \sec^2\left(\frac{\pi}{3}\right) &= 2^2 = 4
\end{align*}

Now substitute into the original expression:

\begin{align*}
2\sin^2\left(\frac{3\pi}{4}\right) + 2\cos^2\left(\frac{\pi}{4}\right) + 2\sec^2\left(\frac{\pi}{3}\right)
&= 2 \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} + 2 \cdot 4 \\
&= 1 + 1 + 8 \\
&= 10
\end{align*}

Hence proved.

Explanation: We have:
\begin{align*}
\sin 75^\circ &= \sin(45^\circ + 30^\circ) \\
&= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ \\
&= \frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2} + \frac{1}{\sqrt{2}} \cdot \frac{1}{2} \\
&= \frac{1}{\sqrt{2}} \cdot \left( \frac{\sqrt{3} + 1}{2} \right) \\
&= \frac{\sqrt{3} + 1}{2\sqrt{2}}
\end{align*}

Explanation: We have:
\begin{align*}
\tan 15^\circ &= \tan(45^\circ - 30^\circ) \\
&= \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \cdot \tan 30^\circ} \\
&= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} \\
&= \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} \\
&= \frac{\sqrt{3} - 1}{\sqrt{3} + 1}
\end{align*}

Now rationalizing the denominator:
\begin{align*}
\tan 15^\circ &= \frac{(\sqrt{3} - 1)^2}{(\sqrt{3} + 1)(\sqrt{3} - 1)} \\
&= \frac{3 + 1 - 2\sqrt{3}}{3 - 1} \\
&= \frac{4 - 2\sqrt{3}}{2} \\
&= 2 - \sqrt{3}
\end{align*}

Explanation:
\begin{align*}
\text{LHS} &= \cos\left( \frac{\pi}{4} - x \right)\cos\left( \frac{\pi}{4} - y \right) - \sin\left( \frac{\pi}{4} - x \right)\sin\left( \frac{\pi}{4} - y \right) \\
&= \cos\left[ \left( \frac{\pi}{4} - x \right) + \left( \frac{\pi}{4} - y \right) \right] \\
&= \cos\left( \frac{\pi}{2} - (x + y) \right) \\
&= \sin(x + y) = \text{RHS} \qquad(\textbf{Hence proved).}
\end{align*}

Explanation: \begin{align*}
\text{LHS} &= \frac{\tan\left( \frac{\pi}{4} + x \right)}{\tan\left( \frac{\pi}{4} - x \right)} \\
&= \frac{ \frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4} \cdot \tan x} }{ \frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4} \cdot \tan x} } \\
&= \frac{ \frac{1 + \tan x}{1 - \tan x} }{ \frac{1 - \tan x}{1 + \tan x} } \\
&= \left( \frac{1 + \tan x}{1 - \tan x} \right)^2 = \text{RHS} \qquad \textbf{Hence proved.}
\end{align*}

Explanation:
\begin{align*}
\text{LHS} &=\frac{\cos(\pi + x)\cos(-x)}{\sin(\pi - x)\cos\left( \frac{\pi}{2} + x \right)}\\&= \frac{(-\cos x)(\cos x)}{(\sin x)(-\sin x)} \\
&= \frac{-\cos^2 x}{-\sin^2 x} \\
&= \frac{\cos^2 x}{\sin^2 x} = \cot^2 x = \text{RHS} \qquad \textbf{Hence proved.}
\end{align*}

Explanation:
\begin{align*}
\text{LHS}&=\cos\left( \frac{3\pi}{2} + x \right)\cos(2\pi + x)\left[ \cot\left( \frac{3\pi}{2} - x \right) + \cot(2\pi + x) \right]\\ &= \sin x \cos x \left[ \tan x + \cot x \right] \\
&= \sin x \cos x \left( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right) \\
&= \sin x \cos x \cdot \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} \\
&= \frac{\sin^2 x + \cos^2 x}{1} = 1 = \text{RHS} \qquad \textbf{Hence proved.}
\end{align*}

Explanation:
\begin{align*}
\text{LHS} &= \sin(n+1)x \cdot \sin(n+2)x + \cos(n+1)x \cdot \cos(n+2)x \\
&= \cos\left[ (n+1)x - (n+2)x \right] \\
&= \cos(-x) = \cos x = \text{RHS} \qquad \textbf{Hence proved.}
\end{align*}

Explanation: We use the identity:


\[
\cos A - \cos B = -2 \sin\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right)
\]



Let:


\[
A = \frac{3\pi}{4} + x, \quad B = \frac{3\pi}{4} - x
\]



Then:
\begin{align*}
\text{LHS} &= \cos\left( \frac{3\pi}{4} + x \right) - \cos\left( \frac{3\pi}{4} - x \right) \\
&= -2 \sin\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right) \\
&= -2 \sin\left( \frac{ \left( \frac{3\pi}{4} + x \right) + \left( \frac{3\pi}{4} - x \right) }{2} \right) \sin\left( \frac{ \left( \frac{3\pi}{4} + x \right) - \left( \frac{3\pi}{4} - x \right) }{2} \right) \\
&= -2 \sin\left( \frac{3\pi}{4} \right) \sin x
\end{align*}

Now:


\[
\sin\left( \frac{3\pi}{4} \right) = \sin\left( \pi - \frac{\pi}{4} \right) = \sin\left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}
\]



So:
\begin{align*}
\text{LHS} &= -2 \cdot \frac{1}{\sqrt{2}} \cdot \sin x \\
&= -\sqrt{2} \sin x = \text{RHS} \qquad \textbf{Hence proved.}
\end{align*}

Explanation: We use the identity:


\[
\sin^2 A - \sin^2 B = (\sin A + \sin B)(\sin A - \sin B)
\]



Let \( A = 6x \), \( B = 4x \). Then:
\begin{align*}
\text{LHS} &= \sin^2(6x) - \sin^2(4x) \\
&= (\sin 6x + \sin 4x)(\sin 6x - \sin 4x)
\end{align*}

Now apply the sum and difference identities:


\[
\sin A + \sin B = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]




\[
\sin A - \sin B = 2 \cos\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right)
\]



So:
\begin{align*}
\sin 6x + \sin 4x &= 2 \sin(5x) \cos(x) \\
\sin 6x - \sin 4x &= 2 \cos(5x) \sin(x)
\end{align*}

Therefore:
\begin{align*}
\text{LHS} &= (2 \sin 5x \cos x)(2 \cos 5x \sin x) \\
&= 4 \sin 5x \cos 5x \sin x \cos x
\end{align*}

Now use the identities:


\[
\sin(2x) = 2 \sin x \cos x, \quad \sin(10x) = 2 \sin 5x \cos 5x
\]



So:


\[
\text{LHS} = \sin(2x) \cdot \sin(10x) = \text{RHS}
\]





\[
\boxed{\sin^2(6x) - \sin^2(4x) = \sin(2x) \cdot \sin(10x)}
\]

Explanation: We use the identity:


\[
\cos^2 A - \cos^2 B = (\cos A + \cos B)(\cos A - \cos B)
\]



Let \( A = 2x \), \( B = 6x \). Then:
\begin{align*}
\text{LHS} &= \cos^2(2x) - \cos^2(6x) \\
&= (\cos 2x + \cos 6x)(\cos 2x - \cos 6x)
\end{align*}

Now apply the sum and difference identities:


\[
\cos A + \cos B = 2 \cos\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]




\[
\cos A - \cos B = -2 \sin\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right)
\]



So:
\begin{align*}
\cos 2x + \cos 6x &= 2 \cos(4x) \cos(-2x) = 2 \cos(4x) \cos(2x) \\
\cos 2x - \cos 6x &= -2 \sin(4x) \sin(-2x) = -2 \sin(4x)(- \sin(2x)) = 2 \sin(4x) \sin(2x)
\end{align*}

Therefore:
\begin{align*}
\text{LHS} &= (2 \cos 4x \cos 2x)(2 \sin 4x \sin 2x) \\
&= 4 \cos 4x \cos 2x \sin 4x \sin 2x
\end{align*}

Now use:


\[
\sin(8x) = 2 \sin(4x) \cos(4x), \quad \sin(4x) = 2 \sin(2x) \cos(2x)
\]



So:


\[
\text{LHS} = \sin(8x) \cdot \sin(4x) = \text{RHS}
\]





\[
\boxed{\cos^2(2x) - \cos^2(6x) = \sin(4x) \cdot \sin(8x)}
\]

Explanation: We begin by grouping terms:
\begin{align*}
\text{LHS} &= \sin 2x + \sin 6x + 2\sin 4x
\end{align*}

Apply the identity:


\[
\sin A + \sin B = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]



Let \( A = 2x \), \( B = 6x \):
\begin{align*}
\sin 2x + \sin 6x &= 2 \sin\left( \frac{2x + 6x}{2} \right) \cos\left( \frac{2x - 6x}{2} \right) \\
&= 2 \sin(4x) \cos(-2x) = 2 \sin(4x) \cos(2x)
\end{align*}

So:
\begin{align*}
\text{LHS} &= 2 \sin(4x) \cos(2x) + 2 \sin(4x) \\
&= 2 \sin(4x) \left( \cos(2x) + 1 \right)
\end{align*}

Now use the identity:


\[
\cos(2x) + 1 = 2 \cos^2 x
\]



Therefore:
\begin{align*}
\text{LHS} &= 2 \sin(4x) \cdot 2 \cos^2 x \\
&= 4 \cos^2 x \cdot \sin(4x) = \text{RHS}
\end{align*}



\[
\boxed{\sin 2x + 2\sin 4x + \sin 6x = 4\cos^2 x \cdot \sin 4x} \qquad
\textbf{Hence proved.}
\]

Explanation: We use the identities:


\[
\sin A + \sin B = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\quad \text{and} \quad
\sin A - \sin B = 2 \cos\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right)
\]


Left-hand side:
\begin{align*}
\cot 4x \cdot (\sin 5x + \sin 3x)
&= \cot 4x \cdot \left[ 2 \sin\left( \frac{5x + 3x}{2} \right) \cos\left( \frac{5x - 3x}{2} \right) \right] \\
&= \cot 4x \cdot \left[ 2 \sin(4x) \cos(x) \right] \\
&= \frac{\cos 4x}{\sin 4x} \cdot 2 \sin(4x) \cos(x) \\
&= 2 \cos 4x \cos x
\end{align*}

Right-hand side:
\begin{align*}
\cot x \cdot (\sin 5x - \sin 3x)
&= \cot x \cdot \left[ 2 \cos\left( \frac{5x + 3x}{2} \right) \sin\left( \frac{5x - 3x}{2} \right) \right] \\
&= \cot x \cdot \left[ 2 \cos(4x) \sin(x) \right] \\
&= \frac{\cos x}{\sin x} \cdot 2 \cos(4x) \sin(x) \\
&= 2 \cos 4x \cos x
\end{align*}

Therefore:


\[
\text{LHS} = \text{RHS}
\quad \Rightarrow \quad
\boxed{\cot 4x \cdot (\sin 5x + \sin 3x) = \cot x \cdot (\sin 5x - \sin 3x)} \qquad \textbf{Hence proved.}
\]

Explanation: We use the identities:


\[
\sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)
\]




\[
\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)
\]



Now, the left-hand side (LHS) becomes:


\[
\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}
= \frac{2 \sin\left(\frac{5x + 3x}{2}\right) \cos\left(\frac{5x - 3x}{2}\right)}{2 \cos\left(\frac{5x + 3x}{2}\right) \cos\left(\frac{5x - 3x}{2}\right)}
\]





\[
= \frac{2 \sin(4x) \cos(x)}{2 \cos(4x) \cos(x)}
= \frac{\sin(4x)}{\cos(4x)} = \tan(4x)
\]



Hence, LHS = RHS. (Proved)

Explanation:
We use the identities:


\[
\sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right)
\]




\[
\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right)
\]



Now, the left-hand side (LHS) becomes:


\[
\frac{\sin x - \sin y}{\cos x + \cos y}
= \frac{2 \cos\left(\frac{x + y}{2}\right) \sin\left(\frac{x - y}{2}\right)}{2 \cos\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right)}
\]





\[
= \frac{\sin\left(\frac{x - y}{2}\right)}{\cos\left(\frac{x - y}{2}\right)} = \tan\left(\frac{x - y}{2}\right)
\]



Hence, LHS = RHS. (Proved)

Explanation: We use the standard trigonometric identities:


\[
\sin A + \sin B = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]




\[
\cos A + \cos B = 2 \cos\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]



Now, consider the left-hand side (LHS):


\[
\frac{\sin x + \sin 3x}{\cos x + \cos 3x}
= \frac{2 \sin\left( \frac{x + 3x}{2} \right) \cos\left( \frac{x - 3x}{2} \right)}{2 \cos\left( \frac{x + 3x}{2} \right) \cos\left( \frac{x - 3x}{2} \right)}
\]





\[
= \frac{2 \sin(2x) \cos(-x)}{2 \cos(2x) \cos(-x)}
= \frac{\sin(2x)}{\cos(2x)} = \tan(2x)
\]



Therefore,


\[
\text{LHS} = \text{RHS}
\quad \boxed{\text{Hence proved.}}
\]

Explanation: We use the following identities:


\[
\sin A - \sin B = 2 \cos\left( \frac{A + B}{2} \right) \sin\left( \frac{A - B}{2} \right)
\]




\[
\cos^2 A - \sin^2 A = \cos 2A \quad \Rightarrow \quad \sin^2 A - \cos^2 A = -\cos 2A
\]



Now, consider the left-hand side (LHS):


\[
\frac{\sin x - \sin 3x}{\sin^2 x - \cos^2 x}
= \frac{2 \cos\left( \frac{x + 3x}{2} \right) \sin\left( \frac{x - 3x}{2} \right)}{-\cos 2x}
\]





\[
= \frac{2 \cos(2x) \sin(-x)}{-\cos 2x}
= \frac{-2 \cos(2x) \sin x}{-\cos 2x}
= 2 \sin x
\]



Therefore,


\[
\text{LHS} = \text{RHS}
\quad \boxed{\text{Hence proved.}}
\]

Explanation: We use the standard identities:


\[
\cos A + \cos B = 2 \cos\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]




\[
\sin A + \sin B = 2 \sin\left( \frac{A + B}{2} \right) \cos\left( \frac{A - B}{2} \right)
\]



Now, consider the left-hand side (LHS):


\[
\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}
= \frac{(\cos 4x + \cos 2x) + \cos 3x}{(\sin 4x + \sin 2x) + \sin 3x}
\]



Apply the sum formulas:


\[
= \frac{2 \cos\left( \frac{4x + 2x}{2} \right) \cos\left( \frac{4x - 2x}{2} \right) + \cos 3x}{2 \sin\left( \frac{4x + 2x}{2} \right) \cos\left( \frac{4x - 2x}{2} \right) + \sin 3x}
\]





\[
= \frac{2 \cos(3x) \cos(x) + \cos(3x)}{2 \sin(3x) \cos(x) + \sin(3x)}
= \frac{\cos(3x)(2 \cos x + 1)}{\sin(3x)(2 \cos x + 1)}
\]





\[
= \frac{\cos(3x)}{\sin(3x)} = \cot(3x)
\]



Therefore,


\[
\text{LHS} = \text{RHS}
\quad \boxed{\text{Hence proved.}}
\]

Explanation: We use the identity:
\[
\cot(A + B) = \frac{\cot A \cot B - 1}{\cot A + \cot B}
\]

Now, consider the left-hand side (LHS):
\begin{align*}
\cot x \cot 2x - \cot 2x \cot 3x - \cot 3x \cot x
&= \cot x \cot 2x - \cot 3x (\cot 2x + \cot x) \\
&= \cot x \cot 2x - \cot(2x + x)(\cot 2x + \cot x) \\
&= \cot x \cot 2x - \left[ \frac{\cot 2x \cot x - 1}{\cot x + \cot 2x} \right](\cot 2x + \cot x) \\
&= \cot x \cot 2x - (\cot 2x \cot x - 1) \\
&= 1 \qquad \textbf{(Hence, proved.)}
\end{align*}

Explanation: We use the identity:
\[
\tan 2A = \frac{2 \tan A}{1 - \tan^2 A}
\]

Now, consider the left-hand side (LHS):
\begin{align*}
\tan 4x &= \tan(2 \cdot 2x) \\
&= \frac{2 \tan 2x}{1 - \tan^2 2x} \\
\text{Using } \tan 2x = \frac{2 \tan x}{1 - \tan^2 x}, \text{ we get:} \\
\tan 4x &= \frac{2 \cdot \left( \frac{2 \tan x}{1 - \tan^2 x} \right)}{1 - \left( \frac{2 \tan x}{1 - \tan^2 x} \right)^2} \\
&= \frac{\frac{4 \tan x}{1 - \tan^2 x}}{1 - \frac{4 \tan^2 x}{(1 - \tan^2 x)^2}} \\
&= \frac{4 \tan x}{1 - \tan^2 x} \cdot \frac{(1 - \tan^2 x)^2}{(1 - \tan^2 x)^2 - 4 \tan^2 x} \\
&= \frac{4 \tan x (1 - \tan^2 x)}{(1 - \tan^2 x)^2 - 4 \tan^2 x} \\
&= \frac{4 \tan x (1 - \tan^2 x)}{1 - 2 \tan^2 x + \tan^4 x - 4 \tan^2 x} \\
&= \frac{4 \tan x (1 - \tan^2 x)}{1 - 6 \tan^2 x + \tan^4 x} \qquad
\textbf{(Hence, proved.)}
\end{align*}

Explanation: We use the following identities:
\begin{align*}
\cos 2A &= 1 - 2 \sin^2 A \\
\sin 2A &= 2 \sin A \cos A
\end{align*}

Now consider the left-hand side:
\begin{align*}
\cos 4x &= \cos(2 \cdot 2x) \\
&= \cos 2(2x) \\
&= 1 - 2 \sin^2(2x) \\
&= 1 - 2 \left( 2 \sin x \cos x \right)^2 \\
&= 1 - 2 \cdot 4 \sin^2 x \cos^2 x \\
&= 1 - 8 \sin^2 x \cos^2 x
\end{align*}

\textbf{Hence, proved.}

Explanation: We use the identity:
\[
\cos 3A = 4 \cos^3 A - 3 \cos A
\]

Also, recall:
\[
\cos 2x = 2 \cos^2 x - 1
\]

Now consider the left-hand side:
\begin{align*}
\cos 6x &= \cos 3(2x) \\
&= 4 \cos^3(2x) - 3 \cos(2x)
\end{align*}

Substitute \(\cos 2x = 2 \cos^2 x - 1\):
\begin{align*}
\cos 6x &= 4 \left( 2 \cos^2 x - 1 \right)^3 - 3 \left( 2 \cos^2 x - 1 \right)
\end{align*}

Now expand the cube:
\[
\left( 2 \cos^2 x - 1 \right)^3 = 8 \cos^6 x - 12 \cos^4 x + 6 \cos^2 x - 1
\]

So:
\begin{align*}
\cos 6x &= 4 \left( 8 \cos^6 x - 12 \cos^4 x + 6 \cos^2 x - 1 \right) - 3 \left( 2 \cos^2 x - 1 \right) \\
&= 32 \cos^6 x - 48 \cos^4 x + 24 \cos^2 x - 4 - 6 \cos^2 x + 3 \\
&= 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1 \qquad \textbf{(Hence, proved.)}
\end{align*}

Explanation: We use the identity:
\[
\cos x + \cos y = 2 \cos\left(\frac{x + y}{2}\right) \cos\left(\frac{x - y}{2}\right)
\]

Now consider the left-hand side:
\begin{align*}
&2\cos\left(\frac{\pi}{13}\right)\cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{3\pi}{13}\right) + \cos\left(\frac{5\pi}{13}\right) \\
&= 2\cos\left(\frac{\pi}{13}\right)\cos\left(\frac{9\pi}{13}\right) + 2\cos\left(\frac{3\pi/13 + 5\pi/13}{2}\right)\cos\left(\frac{3\pi/13 - 5\pi/13}{2}\right) \\
&= 2\cos\left(\frac{\pi}{13}\right)\cos\left(\frac{9\pi}{13}\right) + 2\cos\left(\frac{4\pi}{13}\right)\cos\left(-\frac{\pi}{13}\right) \\
&= 2\cos\left(\frac{\pi}{13}\right)\cos\left(\frac{9\pi}{13}\right) + 2\cos\left(\frac{4\pi}{13}\right)\cos\left(\frac{\pi}{13}\right) \\
&= 2\cos\left(\frac{\pi}{13}\right)\left[\cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{4\pi}{13}\right)\right]
\end{align*}

Now apply the identity again:
\[
\cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{4\pi}{13}\right) = 2\cos\left(\frac{13\pi}{26}\right)\cos\left(\frac{5\pi}{26}\right)
\]

Since \(\cos\left(\frac{13\pi}{26}\right) = \cos\left(\frac{\pi}{2}\right) = 0\), we get:
\[
\cos\left(\frac{9\pi}{13}\right) + \cos\left(\frac{4\pi}{13}\right) = 0
\]

Therefore:
\[
2\cos\left(\frac{\pi}{13}\right) \cdot 0 = 0 \qquad \textbf{Hence, proved.}
\]

Explanation: Step 1: Expand the expression
\begin{align*}
&(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x \\
&= \sin 3x \sin x + \sin^2 x + \cos 3x \cos x - \cos^2 x
\end{align*}

Step 2: Group terms
\[
= (\sin 3x \sin x + \cos 3x \cos x) + (\sin^2 x - \cos^2 x)
\]

Step 3: Use identity
\[
\cos(A - B) = \cos A \cos B + \sin A \sin B
\]
\[
\Rightarrow \sin 3x \sin x + \cos 3x \cos x = \cos(3x - x) = \cos 2x
\]

Also,
\[
\sin^2 x - \cos^2 x = -\cos 2x
\]

Step 4: Combine
\[
\cos 2x + (-\cos 2x) = 0
\]

Hence Proved:
\[
(\sin 3x + \sin x)\sin x + (\cos 3x - \cos x)\cos x = 0
\]

Explanation: \(\textbf{Step 1: }\)Expand the left-hand side
\begin{align*}
&(\cos x + \cos y)^2 + (\sin x - \sin y)^2 \\
&= \cos^2 x + \cos^2 y + 2\cos x \cos y + \sin^2 x + \sin^2 y - 2\sin x \sin y
\end{align*}

\(\textbf{Step 2:}\) Group terms
\[
= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) + 2(\cos x \cos y - \sin x \sin y)
\]

\(\textbf{Step 3:}\) Use identities
\[
\cos^2 x + \sin^2 x = 1,\quad \cos^2 y + \sin^2 y = 1
\]
\[
\Rightarrow \text{LHS} = 1 + 1 + 2(\cos x \cos y - \sin x \sin y) = 2 + 2\cos(x + y)
\]

\(\textbf{Step 4:}\) Use the identity
\[
\cos(x + y) = 2\cos^2\left(\frac{x + y}{2}\right) - 1
\]

\(\textbf{Substitute:}\)
\begin{align*}
\text{LHS} &= 2 + 2\left(2\cos^2\left(\frac{x + y}{2}\right) - 1\right) \\
&= 2 + 4\cos^2\left(\frac{x + y}{2}\right) - 2 \\
&= 4\cos^2\left(\frac{x + y}{2}\right)
\end{align*}

\(\textbf{Therefore,}\)
\[
(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2\left(\frac{x + y}{2}\right)
\]

\(\textbf{Hence Proved.}\)

Explanation: Step 1: Expand the left-hand side
\begin{align*}
&(\cos x - \cos y)^2 + (\sin x - \sin y)^2 \\
&= \cos^2 x + \cos^2 y - 2\cos x \cos y + \sin^2 x + \sin^2 y - 2\sin x \sin y
\end{align*}

Step 2: Group terms
\[
= (\cos^2 x + \sin^2 x) + (\cos^2 y + \sin^2 y) - 2(\cos x \cos y + \sin x \sin y)
\]

Step 3: Use identities
\[
\cos^2 x + \sin^2 x = 1,\quad \cos^2 y + \sin^2 y = 1
\]
\[
\Rightarrow \text{LHS} = 1 + 1 - 2(\cos x \cos y + \sin x \sin y) = 2 - 2\cos(x - y)
\]

Step 4: Use the identity
\[
\cos(x - y) = 1 - 2\sin^2\left(\frac{x - y}{2}\right)
\]

Substitute:
\begin{align*}
\text{LHS} &= 2 - 2\left(1 - 2\sin^2\left(\frac{x - y}{2}\right)\right) \\
&= 2 - 2 + 4\sin^2\left(\frac{x - y}{2}\right) \\
&= 4\sin^2\left(\frac{x - y}{2}\right)
\end{align*}

Therefore,
\[
(\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4\sin^2\left(\frac{x - y}{2}\right) \qquad
\textbf{Hence Proved.}
\]

Explanation: To Prove:
\[
\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x \cos 2x \sin 4x
\]

Step 1: Group terms
\[
\sin x + \sin 3x + \sin 5x + \sin 7x = (\sin x + \sin 5x) + (\sin 3x + \sin 7x)
\]

Step 2: Use the identity
\[
\sin A + \sin B = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]

Apply to each pair:
\begin{align*}
\sin x + \sin 5x &= 2\sin\left(\frac{x + 5x}{2}\right)\cos\left(\frac{x - 5x}{2}\right) = 2\sin(3x)\cos(-2x) \\
\sin 3x + \sin 7x &= 2\sin\left(\frac{3x + 7x}{2}\right)\cos\left(\frac{3x - 7x}{2}\right) = 2\sin(5x)\cos(-2x)
\end{align*}

Step 3: Combine
\[
\text{LHS} = 2\cos(2x)(\sin 3x + \sin 5x)
\]

Apply the identity again:
\[
\sin 3x + \sin 5x = 2\sin\left(\frac{3x + 5x}{2}\right)\cos\left(\frac{3x - 5x}{2}\right) = 2\sin(4x)\cos(-x)
\]

So:
\[
\text{LHS} = 2\cos(2x) \cdot 2\sin(4x)\cos x = 4\cos x \cos 2x \sin 4x
\]

Therefore,
\[
\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x \cos 2x \sin 4x \qquad
\textbf{Hence Proved.}
\]

Explanation: Step 1: Use sum-to-product identities

\[
\sin A + \sin B = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]
\[
\cos A + \cos B = 2\cos\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]

Apply to numerator:
\begin{align*}
\sin 7x + \sin 5x &= 2\sin\left(\frac{7x + 5x}{2}\right)\cos\left(\frac{7x - 5x}{2}\right) = 2\sin(6x)\cos(x) \\
\sin 9x + \sin 3x &= 2\sin\left(\frac{9x + 3x}{2}\right)\cos\left(\frac{9x - 3x}{2}\right) = 2\sin(6x)\cos(3x)
\end{align*}

Apply to denominator:
\begin{align*}
\cos 7x + \cos 5x &= 2\cos(6x)\cos(x) \\
\cos 9x + \cos 3x &= 2\cos(6x)\cos(3x)
\end{align*}

Step 2: Substitute into the expression
\[
\text{LHS} = \frac{2\sin(6x)\cos(x) + 2\sin(6x)\cos(3x)}{2\cos(6x)\cos(x) + 2\cos(6x)\cos(3x)}
\]

Factor numerator and denominator:
\[
= \frac{2\sin(6x)(\cos x + \cos 3x)}{2\cos(6x)(\cos x + \cos 3x)}
\]

Cancel common terms:
\[
= \frac{\sin(6x)}{\cos(6x)} = \tan(6x)
\]

Therefore,
\[
\frac{(\sin 7x + \sin 5x) + (\sin 9x + \sin 3x)}{(\cos 7x + \cos 5x) + (\cos 9x + \cos 3x)} = \tan 6x \qquad \textbf{Hence Proved.}
\]

Explanation: To Prove:
\[
\sin 3x + \sin 2x - \sin x = 4\sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right)
\]

Step 1: Use sum and difference identities

\[
\sin A + \sin B = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]
\[
\sin A - \sin B = 2\sin\left(\frac{A - B}{2}\right)\cos\left(\frac{A + B}{2}\right)
\]

Step 2: Group and simplify
\[
\text{LHS} = \sin 3x + \sin 2x - \sin x
\]

Group \(\sin 2x - \sin x\):
\[
= \sin 3x + 2\sin\left(\frac{2x - x}{2}\right)\cos\left(\frac{2x + x}{2}\right)
= \sin 3x + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right)
\]

Now express \(\sin 3x\) using the identity:
\[
\sin 3x = 2\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right)
\]

So:
\[
\text{LHS} = 2\sin\left(\frac{3x}{2}\right)\cos\left(\frac{3x}{2}\right) + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{3x}{2}\right)
\]

Factor out \(2\cos\left(\frac{3x}{2}\right)\):
\[
= 2\cos\left(\frac{3x}{2}\right)\left[\sin\left(\frac{3x}{2}\right) + \sin\left(\frac{x}{2}\right)\right]
\]

Use the identity again:
\[
\sin A + \sin B = 2\sin\left(\frac{A + B}{2}\right)\cos\left(\frac{A - B}{2}\right)
\]

Apply to the bracket:
\[
= 2\cos\left(\frac{3x}{2}\right) \cdot 2\sin x \cos\left(\frac{x}{2}\right)
= 4\sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right)
\]
Therefore,
\[
\sin 3x + \sin 2x - \sin x = 4\sin x \cos\left(\frac{x}{2}\right) \cos\left(\frac{3x}{2}\right) \qquad \textbf{Hence Proved.}
\]

Explanation: \(\textbf{Step 1: Determine quadrant of \(\frac{x}{2}\):}\)

Since \(x\) lies in the second quadrant, i.e., \(\frac{\pi}{2} < x < \pi\), then:
\[
\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}
\]
So, \(\frac{x}{2}\) lies in the first quadrant. Therefore:
\[
\sin\frac{x}{2}, \cos\frac{x}{2}, \tan\frac{x}{2} \text{ are all positive}
\]

\(\textbf{Step 2: Use identity for \(\sec^2 x\)}\)
\[
\sec^2 x = 1 + \tan^2 x = 1 + \left(-\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{25}{9}
\]
Since \(x\) is in the second quadrant, \(\sec x < 0\), so:
\[
\sec x = -\frac{5}{3} \Rightarrow \cos x = -\frac{3}{5}
\]

\(\textbf{Step 3: Use half-angle identity for cosine}\)
\[
2\cos^2\left(\frac{x}{2}\right) = 1 + \cos x = 1 - \frac{3}{5} = \frac{2}{5}
\Rightarrow \cos^2\left(\frac{x}{2}\right) = \frac{1}{5}
\Rightarrow \cos\left(\frac{x}{2}\right) = \frac{1}{\sqrt{5}}
\]

\(\textbf{Step 4: Use Pythagorean identity}\)
\[
\sin^2\left(\frac{x}{2}\right) = 1 - \cos^2\left(\frac{x}{2}\right) = 1 - \frac{1}{5} = \frac{4}{5}
\Rightarrow \sin\left(\frac{x}{2}\right) = \frac{2}{\sqrt{5}}
\]

\(\textbf{Step 5: Compute tangent}\)
\[
\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2
\]


\(\textbf{Final Answer:}\)
\[
\sin\left(\frac{x}{2}\right) = \frac{2\sqrt{5}}{5}, \quad
\cos\left(\frac{x}{2}\right) = \frac{\sqrt{5}}{5}, \quad
\tan\left(\frac{x}{2}\right) = 2
\]

Explanation: \(\textbf{Step 1: Determine quadrant of \(\frac{x}{2}\)}\)

Since \(x\) lies in the third quadrant, i.e., \(\pi < x < \frac{3\pi}{2}\), then:
\[
\frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}
\]
So, \(\frac{x}{2}\) lies in the second quadrant. Therefore:
\[
\sin\left(\frac{x}{2}\right) > 0,\quad \cos\left(\frac{x}{2}\right) < 0,\quad \tan\left(\frac{x}{2}\right) < 0
\]

\(\textbf{Step 2: Use half-angle identity for cosine}\)
\[
2\cos^2\left(\frac{x}{2}\right) = 1 + \cos x = 1 - \frac{1}{3} = \frac{2}{3}
\Rightarrow \cos^2\left(\frac{x}{2}\right) = \frac{1}{3}
\Rightarrow \cos\left(\frac{x}{2}\right) = -\frac{1}{\sqrt{3}}
\]

\(\textbf{Step 3: Use Pythagorean identity}\)
\[
\sin^2\left(\frac{x}{2}\right) = 1 - \cos^2\left(\frac{x}{2}\right) = 1 - \frac{1}{3} = \frac{2}{3}
\Rightarrow \sin\left(\frac{x}{2}\right) = \sqrt{\frac{2}{3}}
\]

\(\textbf{Step 4: Compute tangent}\)
\[
\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} = \frac{\sqrt{\frac{2}{3}}}{-\frac{1}{\sqrt{3}}} = -\sqrt{2}
\]

\(\textbf{Final Answer:}\)
\[
\sin\left(\frac{x}{2}\right) = \sqrt{\frac{2}{3}}, \quad
\cos\left(\frac{x}{2}\right) = -\frac{1}{\sqrt{3}}, \quad
\tan\left(\frac{x}{2}\right) = -\sqrt{2}
\]

Explanation: \(\textbf{Step 1: Determine quadrant of \(\frac{x}{2}\)}\)

Since \(x\) lies in the second quadrant, i.e., \(\frac{\pi}{2} < x < \pi\), then:
\[
\frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}
\]
So, \(\frac{x}{2}\) lies in the first quadrant. Therefore:
\[
\sin\left(\frac{x}{2}\right) > 0,\quad \cos\left(\frac{x}{2}\right) > 0,\quad \tan\left(\frac{x}{2}\right) > 0
\]

\(\textbf{Step 2: Use Pythagorean identity to find \(\cos x\)}\)

\[
\sin^2 x + \cos^2 x = 1 \Rightarrow \left(\frac{1}{4}\right)^2 + \cos^2 x = 1
\Rightarrow \cos^2 x = 1 - \frac{1}{16} = \frac{15}{16}
\Rightarrow \cos x = -\frac{\sqrt{15}}{4} \quad \text{(since \(x\) is in 2nd quadrant)}
\]

\(\textbf{Step 3: Use half-angle identity for sine}\)

\[
2\sin^2\left(\frac{x}{2}\right) = 1 - \cos x = 1 + \frac{\sqrt{15}}{4}
\Rightarrow \sin^2\left(\frac{x}{2}\right) = \frac{4 + \sqrt{15}}{8}
\Rightarrow \sin\left(\frac{x}{2}\right) = \sqrt{\frac{4 + \sqrt{15}}{8}}
\]

\(\textbf{Step 4: Use Pythagorean identity to find cosine}\)

\[
\cos^2\left(\frac{x}{2}\right) = 1 - \sin^2\left(\frac{x}{2}\right) = 1 - \frac{4 + \sqrt{15}}{8} = \frac{4 - \sqrt{15}}{8}
\Rightarrow \cos\left(\frac{x}{2}\right) = \sqrt{\frac{4 - \sqrt{15}}{8}}
\]

\(\textbf{Step 5: Compute tangent}\)

\[
\tan\left(\frac{x}{2}\right) = \frac{\sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)} =
\frac{\sqrt{\frac{4 + \sqrt{15}}{8}}}{\sqrt{\frac{4 - \sqrt{15}}{8}}} =
\sqrt{\frac{4 + \sqrt{15}}{4 - \sqrt{15}}}
\]

Now rationalize:
\[
\tan\left(\frac{x}{2}\right) = \sqrt{\frac{(4 + \sqrt{15})^2}{(4 - \sqrt{15})(4 + \sqrt{15})}} =
\sqrt{\frac{16 + 8\sqrt{15} + 15}{16 - 15}} =
\sqrt{\frac{31 + 8\sqrt{15}}{1}} = \sqrt{31 + 8\sqrt{15}}
\]

\(\textbf{Final Answer:}\)
\[
\sin\left(\frac{x}{2}\right) = \sqrt{\frac{4 + \sqrt{15}}{8}}, \quad
\cos\left(\frac{x}{2}\right) = \sqrt{\frac{4 - \sqrt{15}}{8}}, \quad
\tan\left(\frac{x}{2}\right) = \sqrt{31 + 8\sqrt{15}}
\]