Question

सिद्ध कीजिए 3. 2sin-1 ⅗ = tan-1 24/7

Answer 1

Answer:

Step-by-step explanation:

माना  $\sin^{-1}(\dfrac{3}{5} )=\theta$  तब  

$\sin\theta=\dfrac{3}{5}$और  $-\dfrac{\pi}{2} \leq \theta\leq \dfrac{\pi}{2}$

∵ $\theta<\dfrac{3}{5} <\dfrac{1}{\sqrt{2} }$, ∴$0<sin\theta<sin\dfrac{\pi}{4}$

$=> 0<\theta<\dfrac{\pi}{4}\\ \\\\=>\cos\theta=\sqrt{1-sin^2\theta} =\sqrt{1-9/25} =\dfrac{4}{5} \\\\\\=>\tan\theta=\dfrac{sin\theta}{cos\theta} =\dfrac{3/5}{4/5}=\dfrac{3}{4}$

$tan2\theta=\dfrac{2\tan\theta}{1-tan^2\theta} =\dfrac{2*\dfrac{3}{4} }{1-(\dfrac{3}{4})^2 } \\\\\\=\dfrac{\dfrac{6}{4} *16}{16-9} =\dfrac{24}{7} \\\\\\=>2\theta= tan^{-1}(\dfrac{24}{7} )$

(∵$0<\theta<\dfrac{\pi}{4} ,0<2\theta<\dfrac{\pi}{2}$)