====== Question 8, Review Exercise ======
Solutions of Question 8 of Review Exercise of Unit 08: Fundamental of Trigonometry. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 8(i)=====
Prove that: $$\sqrt{\frac{\cos \left(90^{\circ}+x\right) \sec (-x) \tan \left(180^{\circ}-x\right)}{\sec \left(360^{\circ}-x\right) \sin \left(180^{\circ}+x\right) \cot \left(90^{\circ}-x\right)}}=i .$$

** Solution. **

\begin{align*}
LHS&= \sqrt{\frac{\cos \left(90^{\circ}+x\right) \sec (-x) \tan \left(180^{\circ}-x\right)}{\sec \left(360^{\circ}-x\right) \sin \left(180^{\circ}+x\right) \cot \left(90^{\circ}-x\right)}}\\
&= \sqrt{\frac{-\sin x  (\sec x) (-\tan x)}{\sec x (- \sin x)\tan x } }\\
&=\sqrt{-1}\\
&=i\\
&=RHS
\end{align*} 

=====Question 8(ii)=====
Prove that: $$\frac{\tan ^{2}\left(\frac{3 \pi}{2}-x\right) \sin ^{2}(\pi+x) \sin (2 \pi-x)}{\cos ^{2}(\pi-x) \cot \left(\frac{3 \pi}{2}+x\right)}=\cos x.$$

** Solution. **

\begin{align*}
LHS &= \frac{\tan ^{2}\left(\frac{3 \pi}{2}-x\right) \sin ^{2}(\pi+x) \sin (2 \pi - x)}{\cos ^{2}(\pi - x) \cot \left(\frac{3 \pi}{2}+x\right)} \\
&= \frac{(\cot x)^{2} (-\sin x)^{2} (-\sin x)}{(-\cos x)^{2} (-\tan x)} \\
&= \frac{\cot ^{2}(x) \sin ^{2}(x) (-\sin x)}{\cos ^{2}(x) (-\tan x)} \\
&= \frac{\frac{\cos^2 x}{\sin^2 x} \cdot \sin^2 x \cdot \sin x}{\cos^2 x \cdot \frac{\sin x}{\cos x}} \\
&= \frac{\cos x}{1} \\
&= \cos x \\
&= RHS
\end{align*}





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