If sinθ = \(\frac{3}4\), prove that \(\sqrt{\frac{cosec^2θ -cot^2θ

1.	If sinθ = \(\frac{3}4\), prove that \(\sqrt{\frac{cosec^2θ -cot^2θ}{sec^2θ-1}}\) = \(\frac{\sqrt{7}}{3}\)
Answer» sin θ = \(\frac{3}4\) ⇒ cosθ = \(\frac{\sqrt{7}}4\) \(\sqrt{\frac{cosec^2θ -cot^2θ}{sec^2θ-1}}\) = \(\sqrt{\frac{1+cot^2θ -cot^2θ}{1+tan^2θ-1}}\) = \(\sqrt{\frac{1}{tan^2\theta}}\) = cotθ = \(\frac{cosθ}{sinθ}\) = \(\frac{\sqrt{7}}4\) = \(\frac{4}3\) = \(\frac{\sqrt{7}}3\)

If sinθ = \(\frac{3}4\), prove that \(\sqrt{\frac{cosec^2θ -cot^2θ}{sec^2θ-1}}\) = \(\frac{\sqrt{7}}{3}\)

Answer»

sin θ = \(\frac{3}4\)

⇒ cosθ = \(\frac{\sqrt{7}}4\)

\(\sqrt{\frac{cosec^2θ -cot^2θ}{sec^2θ-1}}\)

= \(\sqrt{\frac{1+cot^2θ -cot^2θ}{1+tan^2θ-1}}\)

= \(\sqrt{\frac{1}{tan^2\theta}}\)

= cotθ = \(\frac{cosθ}{sinθ}\)

= \(\frac{\sqrt{7}}4\) = \(\frac{4}3\) = \(\frac{\sqrt{7}}3\)