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UPSC NDA Math Model Paper 3

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Question : 101 of 120

Marks: +1, -0

The value of

\tan \left(45+\frac{\theta}{2}\right)

is

tan θ - sec θ
tan θ + sec θ
cot θ - sec θ
cot θ + sec θ

Solution:

Let

\tan \left(45+\frac{\theta}{2}\right)=x

\Rightarrow x=\frac{\tan 45^{\circ}+\tan \frac{\theta}{2}}{1-\tan 45^{\circ}\tan \frac{\theta}{2}}

\Rightarrow x=\frac{1+\tan \frac{\theta}{2}}{1-\tan \frac{\theta}{2}}

(

\because \tan 45^{\circ}=1)

\Rightarrow x=\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}}

\Rightarrow x=\frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}-\sin \frac{\theta}{2}} \times \frac{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}+\sin \frac{\theta}{2}}

\Rightarrow x=\frac{\left(\cos \frac{\theta}{2}+\sin \frac{\theta}{2}\right)^2}{\cos^2 \frac{\theta}{2}-\sin^2 \frac{\theta}{2}}

\Rightarrow x=\frac{\cos^2 \frac{\theta}{2}+\sin^2 \frac{\theta}{2}+2\cos \frac{\theta}{2}\sin \frac{\theta}{2}}{\cos \theta}

\Rightarrow x=\frac{1+\sin \theta}{\cos \theta} \ (\because \sin^2 a+\cos^2 a=1)

\Rightarrow x=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}=\sec \theta+\tan \theta

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