Question

cot²θ-tan²θ=cosec²θ-sec²θ,Prove it

Answer 1

$\bf{LHS}$ = cot²θ - tan²θ
We know from Trigonometric identities ,
cosec²α - cot²α = 1 ⇒cot²α = cosec²α - 1
sec²α - tan²α = 1 ⇒tan²α = sec²α - 1

Hence, LHS = (cosec²θ - 1) - (sec²θ - 1)
= cosec²θ - 1 - sec²θ + 1
= cosec²θ - sec²θ = $\bf{RHS}$

Answer 2

Logic used:

$1) 1+cot^{2}(\theta)= cosec^{2}(\theta) \\ =>cot^{2}(\theta)= cosec^{2}(\theta) -1$ 

$2) 1+tan^{2}(\theta)=sec^{2}(\theta) \\ tan^{2}(\theta)=sec^{2}(\theta)-1$

Steps:

 $1)\: LHS= cot^{2}(\theta)-tan^{2}(\theta)\\ \\ =\ \textgreater \  [cosec^{2}(\theta)-1] - [sec^{2}(\theta)-1] \\ \\ =\ \textgreater \  cosec^{2}(\theta)-1-sec^{2}(\theta))+1 \\ \\ =\ \textgreater \  cosec^{2}(\theta)-sec^{2}(\theta)$ 

Hence Proved :

$cot^{2}(\theta)-tan^{2}(\theta)=cosec^{2}(\theta)-sec^{2}(\theta)$