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.

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*}

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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