Math  /  Trigonometry

QuestionA. -8 B. -2 C. -14 D. -4 E.None Q.6) cos2(tan1x)=\cos ^{2}\left(\tan ^{-1} x\right)= A. x2+1x^{2}+1 B. x2x2+1\frac{x^{2}}{x^{2}+1} C. 1x2+1\frac{1}{x^{2}+1} D. 1x2\frac{1}{x^{2}} E.None

Studdy Solution
Square the cosine to find cos2θ\cos^2 \theta:
cos2θ=(1x2+1)2=1x2+1 \cos^2 \theta = \left(\frac{1}{\sqrt{x^2 + 1}}\right)^2 = \frac{1}{x^2 + 1}
The expression cos2(tan1x)\cos^2(\tan^{-1} x) simplifies to:
1x2+1 \boxed{\frac{1}{x^2 + 1}}
This corresponds to option C.

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