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The value of tan2θ1+tan2θ+cot2θ1+cot2θ\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}+\frac{\cot ^{2} \theta}{1+\cot ^{2} \theta} is equal to

a

0

b

1

c

2

d

3

Answer : Option B
Explanation :

tan2θ1+tan2θ+cot2θ1+cot2θ\frac{\tan ^{2} \theta}{1+\tan ^{2} \theta}+\frac{\cot ^{2} \theta}{1+\cot ^{2} \theta}

=tan2θsec2θ+cot2θcosec2θ=\frac{\tan ^{2} \theta}{\sec ^{2} \theta}+\frac{\cot ^{2} \theta}{\operatorname{cosec}^{2} \theta}

=sin2θcos2θ×cos2θ+cos2θsin2θ×sin2θ=\frac{\sin ^{2} \theta}{\cos ^{2} \theta} \times \cos ^{2} \theta+\frac{\cos ^{2} \theta}{\sin ^{2} \theta} \times \sin ^{2} \theta

=sin2θ+cos2θ=1=\sin ^{2} \theta+\cos ^{2} \theta=1

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