Question

maths experts please solve this question it's very urgent and don't span​

Answer 1

Here I use some identities,

$1.\ \ \dfrac{\sin x}{\cos x}=\tan x\\ \\ \\ 2.\ \ \dfrac{a}{b}-\dfrac{b}{a}=\dfrac{a^2-b^2}{ab}\\ \\ \\ 3.\ \ (a+b)^2-(a-b)^2=4ab\\ \\ \\ 4.\ \ (a-b)(a+b)=a^2-b^2\\ \\ \\ 5.\ \ \tan 2x=\dfrac{2\tan x}{1-\tan^2x}$

So,

$\begin{aligned}&\textsf{LHS}\\ \\ \Longrightarrow\ \ &\dfrac{\cos x+\sin x}{\cos x-\sin x}-\dfrac{\cos x-\sin x}{\cos x+\sin x}\\ \\ \Longrightarrow\ \ &\dfrac{\frac{\cos x+\sin x}{\cos x}}{\frac{\cos x-\sin x}{\cos x}}-\dfrac{\frac{\cos x-\sin x}{\cos x}}{\frac{\cos x+\sin x}{\cos x}}\\ \\ \Longrightarrow\ \ &\frac{1+\tan x}{1-\tan x}-\frac{1-\tan x}{1+\tan x}\\ \\ \Longrightarrow\ \ &\frac{(1+\tan x)^2-(1-\tan x)^2}{(1-\tan x)(1+\tan x)}\\ \\ \Longrightarrow\ \ &\frac{4\tan x}{1-\tan^2x}\end{aligned}$

$\Longrightarrow\ \ 2\cdot \dfrac{2\tan x}{1-\tan^2x}\\ \\ \\ \Longrightarrow\ \ 2\tan 2x\\ \\ \\ \Longrightarrow\ \ \textsf{RHS}$

Hence Proved!

Answer 2

Answer:

please find the attachment

may be it's helpful for you ✴️✴️