RRB Junior Engineer Stage 1 Model Papers 10

© examsiri.com

Question : 9 of 98

Marks: +1, -0

If

\tan α = n \tan β

, and

\sin α = m \sin β

, then

\cos^2 α

is

${m^2-1}/{n^2-1}$
${m^2+1}/{n^2+1}$
${m^2}/{n^2+1}$
${m^2}/{n^2}$

Solution:

\tan α = n \tan β

⇒\tan β=1/n \tan α

⇒\cot β=1/{\tan α}

and

\sin α=m \sin β⇒\sin β=1/m \sin α

⇒\cosec β=m/{\sin α}

∴\cosec^2 β-\cot^2 β=1

⇒m^2/{\sin^2 α}-n^2/{\tan^2 α}=1

=m^2/{\sin2 α}-{n^2 \cos^2α}/{\sin^2α}=1

⇒{m^2-n^2\cos^2α}/{\sin^2 α}=1

⇒m^2-n^2\cos^2 α=\sin^2 α

=1-cos^2 α

⇒m^2-1=n^2\cos^2 α-cos^2 α

=(n^2-1)\cos^2 α

=\cos^2 α={m^2-1}/{n^2-1}

© examsiri.com

Go to Question:

Prev Question

Next Question