Question

prove that tan^2 theta - sin^2 theta = tan^2 theta * 1/cosec^2 theta

Answer 1

Solution:

${tan}^{2} \theta - {sin}^{2} \theta \\ \\ \frac{ {sin}^{2} \theta}{ {cos}^{2} \theta} - {sin}^{2} \theta \\ \\ = > take \: {sin}^{2} \theta \: common \\ \\ {sin}^{2} \theta( \frac{1}{{cos}^{2} \theta} - 1) \\ \\ {sin}^{2} \theta( \frac{1 - {cos}^{2} \theta}{ {cos}^{2}\theta } ) \\ \\ = {sin}^{2} \theta( \frac{{sin}^{2} \theta}{{cos}^{2} \theta} ) \\ \\ since \: 1 - {cos}^{2} \theta = {sin}^{2} \theta \\ \\ so \\ \\ \frac{ {sin}^{2}\theta }{ {cos}^{2} \theta} \times {sin}^{2}\theta \\ \\ = {tan}^{2}\theta \times \frac{1}{ {cosec}^{2}\theta } \\ \\hence \: proved \\ \\ because \\ \\ {sin}^{2} \theta = \frac{1}{ {cosec}^{2}\theta }$
Hope it helps you

Answer 2

To Prove ⇒ tan²θ - sin²θ = tan²θ ×  1/cosec²θ

Proof ⇒

L.H.S. = tan²θ - sin²θ

= Sin²θ/Cos²θ - Sin²θ          [tan²θ =  Sin²θ/Cos²θ]

= Sin²θ(1/Cos²θ - 1)

= Sin²θ/Cos²θ(1 - Cos²θ)

= Sin²θ × Sin²θ/Cos²θ      [Sin²θ = 1 - Cos²θ]

_________________________________

R.H.S. = tan²θ × 1/Cosec²θ

= tan²θ × Sin²θ      [Sin²θ   = 1/Cosec²θ]

= Sin²θ/Cos²θ × Sin²θ  

= L.H.S.

Hence Proved.

Hope it helps.