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अगर 5tanθ=5sinθ\sqrt{5} \tan \theta=5 \sin \theta, फिर (sin2θ\left(\sin ^{2} \theta\right. cos2θ)?\left.-\cos ^{2} \theta\right ) ? का मूल्य क्या है

a

35\frac{3}{5}

b

15\frac{1}{5}

c

45\frac{4}{5}

d

25\frac{2}{5}

Answer : Option A
Explanation :

5tanθ=5sinθ\sqrt{5} \tan \theta=5 \sin \theta

5sinθcosθ=5sinθ\Rightarrow \sqrt{5} \frac{\sin \theta}{\cos \theta}=5 \sin \theta

5sinθcosθ5sinθ=0\Rightarrow \frac{\sqrt{5} \sin \theta}{\cos \theta}-5 \sin \theta=0

5sinθ(1cosθ5)=0\Rightarrow \sqrt{5} \sin \theta\left(\frac{1}{\cos \theta}-\sqrt{5}\right)=0

5sinθ(1cosθ5)=0\Rightarrow \sqrt{5} \sin \theta\left(\frac{1}{\cos \theta}-\sqrt{5}\right)=0

cosθ=15\Rightarrow \cos \theta=\frac{1}{\sqrt{5}} or sinθ=0\sin \theta=0

sinθ=1cos2θ=115=45=25\therefore \sin \theta=\sqrt{1-\cos ^{2} \theta}=\sqrt{1-\frac{1}{5}}=\sqrt{\frac{4}{5}}=\frac{2}{\sqrt{5}}

sin2θcos2θ=4515=35\therefore \sin ^{2} \theta-\cos ^{2} \theta=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}

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