Question

(tan theta + cot theta + sec theta )×(tan theta +cot theta -sec theta) =cosec^2 theta. Prove it​

Answer 1

Answer:-

(tanθ+cotθ+secθ)×(tanθ+cotθ-secθ)=$cosec^{2}θ$

Step-by-step explanation:

To proof-(tanθ+cotθ+secθ)(tanθ+cotθ-secθ)=$cosec^{2}\theta$

Taking Left Hand Side i.e L.H.S

=(tanθ+cotθ+secθ)(tanθ+cotθ-secθ)

=tanθ(tanθ+cotθ-secθ)+cotθ(tanθ+cotθ-secθ)+secθ(tanθ+cotθ-secθ)

$=tan^{2}\theta$+tanθcotθ-tanθsecθ+cotθtanθ-cotθsecθ+$cot^{2}\theta$+secθtanθ+secθcotθ-$sec^{2}\theta$

$=tan^{2}\theta+tan\theta\times \frac{1}{tan\theta}-$tanθsecθ+$cot\theta\times \frac{1}{cot\theta}+cot^{2}\theta$-cotθsecθ+secθtanθ+secθcotθ-$sec^{2}\theta$

$Using\ 1+tan^{2}\theta=sec^{2}\theta$

$Using\ 1+cot^{2}\theta=cosec^\theta$

=$1+tan^{2}\theta$-tanθsecθ+1+$cot^{2}\theta$-cotθsecθ+secθtanθ+secθcotθ-$sec^{2}\theta$

=$sec^{2}\theta$-tanθsecθ+$cosec^{2}\theta$-cotθsecθ+secθtanθ+secθcotθ-$sec^{2}\theta$

=$cosec^{2}\theta$

=R.H.S

Hence L.H.S=R.H.S