SSC CGL Tier II 25 Oct 2015 Paper 1

Question : 97 of 100

Marks: +1, -0

If tan A = n tan B and sin A = m sin B, then the value of

\cos^2A

Solution:

(4) tan A = n tan B

⇒\tan B= {1}/{n} \tan A

⇒ \cot B= {n}/{\tan A}

and,

\sin A=m \sin B

⇒ \sin B= {1}/{m} \sin A

⇒ \cosec B= {m}/{\sin A}

∵ \cosec ^{2} B-\cot ^{2} B=1

⇒{m^{2}}/{\sin ^{2} A}- {n^{2}}/{\tan ^{2} A}=1

⇒{m^{2}}/{\sin ^{2} A}- {n^{2} \cos ^{2} A}/{\sin ^{2} A}=1

⇒{m^{2}-n^{2} \cos ^{2} A}/{\sin ^{2} A}=1

⇒m^{2}-n^{2} \cos ^{2} A=\sin ^{2} A

⇒ m^{2}-n^{2} \cos ^{2} A=1-\cos ^{2} A

⇒ m^{2}-1=n^{2} \cos ^{2} A-\cos ^{2} A

⇒ m^{2}-1= (n^{2}-1 ) \cos ^{2} A

⇒ \cos ^{2} A= {m^{2}-1}/{n^{2}-1}

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