Question 8(i, ii & iii) Exercise 8.2

Solutions of Question 8(i, ii & iii) of Exercise 8.2 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.

Verify the identities: $(\sin \theta+\cos \theta)^{2}=1+\sin 2 \theta$

Solution. $$\begin{align*} LHS & = (\sin \theta+\cos \theta)^{2} \\ &=\sin^2\theta + \cos^2\theta +2\sin \theta \cos\theta\\ &= 1+2\sin \theta \cos\theta \quad (\because \sin^2\theta+\cos^2\theta=1) \\ &=1+\sin 2\theta \quad (\because \sin2\theta=2\sin\theta \cos\theta) \\ &=RHS \end{align*}$$

Verify the identities: $\tan 2 x=\dfrac{1}{1-\tan x}-\dfrac{1}{1+\tan x}$

Solution.

$$\begin{align*} RHS & = \frac{1}{1-\tan x}-\frac{1}{1+\tan x} \\ &= \frac{1+\tan x-1+\tan x}{(1-\tan x)(1+\tan x)} \\ &= \frac{2\tan x}{1-\tan^2 x} \\ &=\tan 2x \\ & = LHS \end{align*}$$

Verify the identities: $\tan \frac{\theta}{2}=\frac{\sin \theta}{1+\cos \theta}$

Solution.

$$\begin{align*} RHS &=\frac{\sin \theta}{1+\cos \theta}\\ &= \frac{2\sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2\cos^2 \frac{\theta}{2}}\quad \text(by \,using\, half\, angle\, identity)\\ &=\tan \frac{\theta}{2}\\ &= RHS \end{align*}$$

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