Question

50/cosec^2 theta +50 /sec^2 theta​

Answer 1

ANSWER:

To Find:

• Value of 50/cosec^2 theta +50 /sec^2 theta

Solution:

$\text{We are given that,}\\\\:\longrightarrow \dfrac{50}{\csc^2\theta}+\dfrac{50}{\sec^2\theta}\\\\\text{We know that,}\\\\:\hookrightarrow\csc\theta=\dfrac{1}{\sin\theta}\\\\\text{And,}\\\\:\hookrightarrow\sec\theta=\dfrac{1}{\cos\theta}\\\\\text{So,}\\\\:\implies \dfrac{50}{\csc^2\theta}+\dfrac{50}{\sec^2\theta}\\\\:\implies\dfrac{50}{\dfrac{1}{\sin^2\theta}}+\dfrac{50}{\dfrac{1}{\cos^2\theta}}\\\\:\implies 50(\sin^2\theta)+50(\cos^2\theta)\\\\\text{Taking 50 common,}\\\\:\implies 50(\sin^2\theta+\cos^2\theta)\\\\\text{We know that,}\\\\:\hookrightarrow \sin^2\theta+\cos^2\theta=1\\\\\text{So,}\\\\:\implies50(\sin^2\theta+\cos^2\theta)\\\\:\implies50(1)\\\\\bf{:\implies \dfrac{50}{\csc^2\theta}+\dfrac{50}{\sec^2\theta}=50}$

Formulae Used:

$:\hookrightarrow 1)\:\csc\theta=\dfrac{1}{\sin\theta}\\\\:\hookrightarrow2)\:\sec\theta=\dfrac{1}{\cos\theta}\\\\:\hookrightarrow3)\:\sin^2\theta+\cos^2\theta=1$