Question

solve this question please ???​

Answer 1

Answer:

HEY FRIEND HERE'S YOUR ANSWER

Step-by-step explanation:

PLEASE MARK ME AS BRAINLIEST

Answer 2

Given LHS :

• $\sf{(1+cot\:\theta\:+\:cosec\:\theta\:)(1+cot\:\theta\:-\:cosec\:\theta)}$

Given RHS :

• $\sf{2cot\:\theta}$

To Prove :

• $\sf{(1+cot\:\theta\:+\:cosec\:\theta\:)(1+cot\:\theta\:-\:cosec\:\theta)\:=\:2\:cot\:\theta}$

Proof :

Taking LHS,

$\sf{(1+cot\:\theta\:+\:cosec\:\theta\:)(1+cot\:\theta\:-\:cosec\:\theta)}$

$\sf{\Big(1\:+\:\dfrac{cos\:\theta}{sin\:\theta}\:+\:\dfrac{1}{sin\:\theta}\Big)\:\Big(1\:+\:\dfrac{cos\:\theta}{sin\:\theta}\:-\:\dfrac{1}{sin\:\theta}\Big)}$

$\sf{\Big[\because\:cot\:\theta\:=\:\dfrac{cos\:\theta}{sin\:\theta}\:\:;\:\:cosec\:\theta\:=\:\dfrac{1}{sin\:\theta}\Big]}$

$\sf{\big(\:1\:+\:\dfrac{cos\:\theta\:+\:1}{Sin\:\theta}\big)\:\big(\:1\:+\:\dfrac{cos\:\theta\:-\:1}{sin\:\theta}\big)}$

$\sf{\big(\dfrac{sin\:\theta\:+\:cos\:\theta\:+\:1}{sin\:\theta}\big)\:\big(\dfrac{sin\:\theta\:+\:cos\:\theta\:-\:1}{sin\:\theta}\big)}$

$\sf{\dfrac{(sin\:\theta\:+\:cos\:\theta)^2\:-\:(1)^2}{sin\:^2\:\theta}}$

$\sf{\dfrac{sin^2\:\theta\:+\:2\:sin\:\theta\:\times\:cos\:\theta\:-\:1}{sin^2\:\theta}}$

$\sf{\big[\because\:(a+b)^2\:=\:a^2\:+\:2ab\:+\:b^2)\:\big]}$

$\sf{\dfrac{sin^2\:\theta\:+\:cos^2\:\theta\:+\:2\:sin\:\theta\:\times\:cos\:\theta\:-\:1}{sin^2\:\theta}}$

$\sf{\dfrac{1\:+\:2\:sin\:\theta\:\times\:cos\:\theta\:-\:1}{sin^2\:\theta}}$

$\sf{\big[\because\:sin^2\:\theta\:+\:cos^2\:\theta\:=\:1\:\big]}$

$\sf{\dfrac{2\:sin\:\theta\:\times\:cos\:\theta}{sin^2\:\theta}}$

$\sf{\dfrac{2\:\cos\:\theta\:}{sin\:\theta}}$

$\sf{2\:cot\:\theta}$

$\large{\boxed{\bold{(1+cot\:\theta\:+\:cosec\:\theta\:)(1+cot\:\theta\:-\:cosec\:\theta\:=\:2\:cot\:\theta}}}$