SSC CGL Tier II 1 Dec 2016 Paper 1

Question : 88 of 100

Marks: +1, -0

If secθ + tanθ = m (>1), then the value of sinθ is (0° < θ < 90°)

Solution:

(2)

\sec \θ+\tan \θ=m .(

Given

)...(i)

\∵ \sec ^{2} \θ-\tan ^{2} \θ=1

⇒(\sec \θ+\tan \θ)(\sec \θ-\tan \θ)

=1

⇒ \sec \θ-\tan \θ= {1}/{m}....(ii)

By equations

(i) + (ii),

2 \sec \θ=m+ {1}/{m}

⇒ \sec \θ= {m^{2}+1}/{2 m}

By equation

(i) – (ii)

2 \tan \θ=m- {1}/{m}

⇒ \tan \θ= {m^{2}-1}/{2 m}

∴ \sin \θ= {\tan \θ}/{\sec \θ}

= {m^{2}-1}/{2 m}×{2 m}/{m^{2}+1}= {m^{2}-1}/{m^{2}+1}

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