Question 9, Review Exercise

Solutions of Question 9 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.

Simplify: $$\sqrt{\frac{\left(1-\tan ^{2} x \cos (-x) \cos \left(360^{\circ}-x\right)\right) \tan 45^{\circ}}{\left\{\sin 90^{\circ}-\sin \left(180^{\circ}+x\right)\right\}\left\{\sin 90^{\circ}-\cos \left(90^{\circ}-x\right)\right\}}}$$

Solution.

$$\begin{align*} & \sqrt{\frac{\left(1-\tan ^{2} x \cos (-x) \cos \left(360^{\circ}-x\right)\right) \tan 45^{\circ}}{\left\{\sin 90^{\circ}-\sin \left(180^{\circ}+x\right)\right\}\left\{\sin 90^{\circ}-\cos \left(90^{\circ}-x\right)\right\}}} \\ &= \sqrt{\frac{\left(1 - \tan^2 x \cdot \cos x \cdot \cos x\right) \cdot 1}{\left\{1 - (-\sin x)\right\} \left\{1 - \sin x\right\}}} \\ &= \sqrt{\frac{1 - \tan^2 x \cdot \cos^2 x}{(1 + \sin x)(1 - \sin x)}} \\ &= \sqrt{\frac{1 - \frac{\sin^2 x}{\cos^2 x} \cdot \cos^2 x}{1 - \sin^2 x}} \\ &= \sqrt{\frac{1 - \sin^2 x}{\cos^2 x}} \\ &= \sqrt{\frac{\cos^2 x}{\cos^2 x}} \\ &= 1 \end{align*}$$

< Question 8 Question 10 >