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UPSC CDS 2 2020 Math Paper

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Question : 85 of 100

Marks: +1, -0

If

\csc \theta - \sin \theta = m

\sec \theta - \cos \theta = n

m^{\frac{4}{3}} n^{\frac{2}{3}} + m^{\frac{2}{3}} n^{\frac{4}{3}}

equal to ?

0
1
$mn$
$m^2 n^2$

Solution:

we have

\cos \theta - \sin \theta = m

and

\sec \theta - \cos \theta = n

\frac{1}{\sin \theta} - \sin \theta = m

\frac{1 - \sin^2 \theta}{\sin \theta} = m

\frac{\cos^2 \theta}{\sin \theta} = m

......(i)

\frac{1}{\cos \theta} - \cos \theta = n

\frac{1 - \cos^2 \theta}{\cos \theta} = n

\frac{\sin^2 \theta}{\cos \theta} = n

......(ii)

m^2 n = \frac{\cos^4 \theta}{\sin^2 \theta} \cdot \frac{\sin^2 \theta}{\cos \theta} = \cos^3 \theta,

m n^2 = \frac{\cos^2 \theta}{\sin \theta} \cdot \frac{\sin^4 \theta}{\cos^2 \theta} = \sin^3 \theta

m^{\frac{4}{3}} n^{\frac{2}{3}} + m^{\frac{2}{3}} n^{\frac{4}{3}}

= (m^2 n)^{\frac{2}{3}} + (m n^2)^{\frac{2}{3}}

= (\cos^3 \theta)^{\frac{2}{3}} + (\sin^3 \theta)^{\frac{2}{3}}

= \cos^2 \theta + \sin^2 \theta

= 1

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