SSC CHSL Model Paper 19

Question : 64 of 99

Marks: +1, -0

Find the value of tan θ(1 + sec 2θ) (1 + sec 4θ) (1 + sec8θ).

Solution:

\tan θ(1 + \sec 2θ) (1 + \sec 4θ) (1 + \sec8θ)

=\tanθ(1+1/{\cos2θ})(1+1/{\cos4θ})

(1+1/{\cos8θ})

=\tanθ( {\cos2θ +1}/{\cos2θ})( {\cos4θ +1}/{\cos4θ})

( {\cos8θ +1}/{\cos8θ})

=\tanθ\ {2\cos^2θ}/{\cos 2θ}. {2\cos^2 2θ}/{\cos 4θ}. {2\cos^2 4θ}/{\cos 8θ}

[ ∵ 1 + \cos2θ = 2 \cos^2θ]

8. {\tan θ.\cos2 θ.\cos 2θ. \cos 4θ}/{\cos 8θ}

4.{2 \sin θ \cos θ. \cos 2θ. \cos 4θ}/{\cos 8θ}

[2\sinθ.\cosθ = \sin20]

4.{\sin 2θ. \cos 2θ. \cos 4θ}/{\cos 8θ}

{2 \sin 4θ. \cos 4θ}/{\cos 8θ}={\sin8θ}/{\cos 8θ}

=\tan 8θ

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