Question

cot theta -tan theta over 1 - 2 sin square theta is equal to cosec theta into sec theta​

Answer 1

$\mathsf{Given :\;\dfrac{cot\theta - tan\theta}{1 - 2sin^2\theta}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{cos^2\theta + sin^2\theta = 1}}}$

$\mathsf{Substitute\;(cos^2\theta + sin^2\theta)\;in\;the\;place\;of\;1\;in\;denominator}$

$\mathsf{\implies \dfrac{cot\theta - tan\theta}{cos^2\theta + sin^2\theta - 2sin^2\theta}}$

$\mathsf{\implies \dfrac{cot\theta - tan\theta}{cos^2\theta - sin^2\theta}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{cot\theta = \dfrac{cos\theta}{sin\theta}}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{tan\theta = \dfrac{sin\theta}{cos\theta}}}}$

$\mathsf{\implies \dfrac{\dfrac{cos\theta}{sin\theta} - \dfrac{sin\theta}{cos\theta}}{cos^2\theta - sin^2\theta}}$

Take LCM among the fractions in the Numerator :

$\mathsf{\implies \dfrac{\dfrac{cos^2\theta - sin^2\theta}{sin\theta.cos\theta}}{cos^2\theta - sin^2\theta}}$

$\mathsf{\implies \dfrac{cos^2\theta - sin^2\theta}{sin\theta.cos\theta(cos^2\theta - sin^2\theta)}}$

$\mathsf{\implies \dfrac{1}{sin\theta.cos\theta}}$

$\mathsf{\implies \bigg(\dfrac{1}{sin\theta}\bigg)\bigg(\dfrac{1}{cos\theta}\bigg)}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{co{sec}\theta = \dfrac{1}{sin\theta}}}}$

$\bigstar\;\;\textsf{We know that : \boxed{\mathtt{{sec}\theta = \dfrac{1}{cos\theta}}}}$

$\implies \large\boxed{\mathsf{co{sec}\theta.{sec}\theta}}$

Answer 2

Step-by-step explanation:

its your answer.. hope it will be helpful for you ..