Question

under root 1 + sin square theta sec square theta / 1+cos square theta cosec square theta is equal to tan theta​

Answer 1

$\underline{\textsf{To prove:}}$

$\mathsf{\displaystyle\sqrt{\dfrac{1+sin^2\theta\,sec^2\theta}{1+cos^2\theta\,cosec^2\theta}}=tan\,\theta}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\displaystyle\sqrt{\dfrac{1+sin^2\theta\,sec^2\theta}{1+cos^2\theta\,cosec^2\theta}}}$

$\mathsf{=\displaystyle\sqrt{\dfrac{1+sin^2\theta(\frac{1}{cos^2\theta})}{1+cos^2\theta(\frac{1}{sin^2\theta})}}}$

$\mathsf{=\displaystyle\sqrt{\dfrac{\dfrac{cos^2\theta+sin^2\theta}{cos^2\theta}}{\dfrac{sin^2\theta+cos^2\theta}{sin^2\theta}}}}$

$\textsf{Using}$

$\boxed{\mathsf{sin^2A+cos^2A=1}}$

$\mathsf{=\displaystyle\sqrt{\dfrac{\dfrac{1}{cos^2\theta}}{\dfrac{1}{sin^2\theta}}}}$

$\mathsf{=\displaystyle\sqrt{\dfrac{1}{cos^2\theta}{\times}\dfrac{sin^2\theta}{1}}}$

$\mathsf{=\displaystyle\sqrt{\dfrac{sin^2\theta}{cos^2\theta}}}$

$\mathsf{=\displaystyle\sqrt{tan^2\theta}}$

$\mathsf{=tan\theta}$

$\implies\boxed{\mathsf{\displaystyle\sqrt{\dfrac{1+sin^2\theta\,sec^2\theta}{1+cos^2\theta\,cosec^2\theta}}=tan\,\theta}}$

Answer 2

Step-by-step explanation:

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