Find the value of $\tan 22\Upsilon30^{\prime}$.
Let $\quad \theta=45 \Upsilon$
We know that, $\quad \tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{\theta}{2}}{2 \cos ^2 \frac{\theta}{2}} \Rightarrow \tan \frac{\theta}{2}=\frac{\sin \theta}{1+\cos \theta}$
$$\begin{aligned} \therefore \quad \tan 22 \Upsilon 30^{\prime} & =\frac{\sin 45 \Upsilon}{1+\cos 45 \Upsilon} & {[\because \theta=45 \Upsilon] } \\ & =\frac{\frac{1}{\sqrt{2}}}{1+\frac{1}{\sqrt{2}}}=\frac{1}{\sqrt{2}+1} & \end{aligned}$$
Prove that $\sin 4 A=4 \sin A \cos ^3 A-4 \cos A \sin ^3 A$.
$\begin{aligned} \text { LHS } & =\sin 4 A \\ & =2 \sin 2 A \cdot \cos 2 A \\ & =2(2 \sin A \cdot \cos A)\left(\cos ^2 A-\sin ^2 A\right) \quad \left[\begin{array}{l} \because \cos 2 A=\cos ^2 A-\sin ^2 A \\ \text { and } \sin 2 A=2 \sin A \cdot \cos A \end{array}\right]\\ & =4 \sin A \cdot \cos ^3 A-4 \cos A \sin ^3 A \\ \therefore \quad \text { LHS } & =\text { RHS }\quad \text{Hence proved.}\end{aligned}$
If $\tan \theta+\sin \theta=m$ and $\tan \theta-\sin \theta=n, \quad$ then prove that $m^2-n^2=4 \sin \theta \tan \theta$
Given that, $\tan\theta+\sin\theta=m\quad \text{... (i)}$
and $\tan\theta-\sin\theta=n\quad \text{... (ii)}$
Now, $\begin{aligned} & m+n=\tan \theta+\sin \theta+\tan \theta-\sin \theta \\ & m+n=2 \tan \theta\quad \text{... (iii)}\end{aligned}$
Also, $\begin{aligned} & m-n=\tan \theta+\sin \theta-\tan \theta+\sin \theta \\ & m-n=2 \sin \theta\quad \text{... (iv)}\end{aligned}$
From Eqs. (iii) and (iv),
$$\begin{aligned} (m+n)(m-n) & =4 \sin \theta \cdot \tan \theta \\ m^2-n^2 & =4 \sin \theta \cdot \tan \theta\quad \text{Hence proved} \end{aligned}$$
If $\tan (A+B)=p$ and $\tan (A-B)=q$, then show that $\tan 2 A=\frac{p+q}{1-p q}$.
Given that $\tan (A+B)=p\quad \text{... (i)}$
and $\tan (A-B)=q\quad \text{.... (ii)}$
$\begin{aligned} \therefore \quad \tan 2 A & =\tan (A+B+A-B) \\ & =\frac{\tan (A+B)+\tan (A-B)}{1-\tan (A+B) \tan (A-B)} \quad\left[\because \tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}\right] \\ & =\frac{p+q}{1-p q} \quad \text { [from Eqs. (i) and (ii)] }\end{aligned}$
If $\cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta$, then prove that $\cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta)$
$\begin{aligned} & \text { Given that, } \quad \cos \alpha+\cos \beta=0=\sin \alpha+\sin \beta \\ & \Rightarrow \quad(\cos \alpha+\cos \beta)^2-(\sin \alpha+\sin \beta)^2=0 \\ & \Rightarrow \quad \cos ^2 \alpha+\cos ^2 \beta+2 \cos \alpha \cos \beta-\sin ^2 \alpha-\sin ^2 \beta-2 \sin \alpha \sin \beta=0 \\ & \Rightarrow \quad \cos ^2 \alpha-\sin ^2 \alpha+\cos ^2 \beta-\sin ^2 \beta=2(\sin \alpha \sin \beta-\cos \alpha \cos \beta) \\ & \Rightarrow \quad \cos 2 \alpha+\cos 2 \beta=-2 \cos (\alpha+\beta)\quad \text{Hence proved.}\end{aligned}$