Question 2 and 3, Review Exercise

Solutions of Question 2 and 3 of Review Exercise of Unit 09: Trigonometric Functions. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

If $\cos \theta -\sin \theta=\sqrt{2}\sin \theta,$ then show that $\cos \theta+ \sin \theta=\sqrt{2} \cos \theta$

Solution.

Given $$\cos \theta -\sin \theta=\sqrt{2}\sin \theta$$ This gives $$\begin{align*} & \cos \theta=\sqrt{2}\sin \theta + \sin \theta \\ \implies & \cos \theta=(\sqrt{2}+1)\sin \theta\\ \implies & \sin \theta=\frac{1}{\sqrt{2}+1}\cos \theta ... (1) \end{align*}$$ Now $$\begin{align*} LHS & = \cos \theta+ \sin \theta \\ & = \cos \theta+ \frac{1}{\sqrt{2}+1}\cos \theta \quad \text{from (1)}\\ & = \left(1+ \frac{1}{\sqrt{2}+1}\right)\cos \theta \\ & = \frac{\sqrt{2}+1+1}{\sqrt{2}+1} \cos \theta \\ & = \frac{2+\sqrt{2}}{\sqrt{2}+1} \cos \theta \\ & = \sqrt{2} \left(\frac{\sqrt{2}+1}{\sqrt{2}+1}\right) \cos \theta \\ & = \sqrt{2} \cos \theta \\ & = RHS \end{align*}$$

Verify: $\dfrac{\tan x - \cot x}{\sin x \cos x} = \sec^2 x - \csc^2 x$

Solution. $$\begin{align*} LHS & = \dfrac{\tan x - \cot x}{\sin x \cos x} \\ & = \dfrac{\tan x}{\sin x \cos x} - \dfrac{\cot x}{\sin x \cos x} \\ & = \dfrac{\sin x}{\cos x \sin x \cos x} - \dfrac{\cos x}{\sin x \sin x \cos x} \\ & = \dfrac{1}{\cos^2} - \dfrac{1}{\sin^2 x} \\ & = \sec^2 x - \csc^2 x \\ & = RHS \end{align*}$$

Verify: $\dfrac{\sec^4 x - \tan^4 x}{\sec^2 x + \tan^2 x} = \sec^2 x - \tan^2 x$

Solution.

$$\begin{align*} LHS & = \dfrac{\sec^4 x - \tan^4 x}{\sec^2 x + \tan^2 x} \\ & = \dfrac{(\sec^2 x)^2 - (\tan^2 x)^2}{\sec^2 x + \tan^2 x} \\ & = \dfrac{(\sec^2 x + \tan^2 x)(\sec^2 x - \tan^2 x)}{\sec^2 x + \tan^2 x} \\ & = \sec^2 x - \tan^2 x \\ & = RHS \end{align*}$$

Verify: $\dfrac{\sin t}{1- \cos t}-\dfrac{\sin t \cos t}{1+ \cos t} = \csc (1+\cos^2 t)$

Solution.

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