If \sin x-\cos x=0,0^\circ

SSC CHSL 15 Oct 2020 Shift 1 Paper

Question : 69 of 100

Marks: +1, -0

\sin x-\cos x=0,0° < x < 90°

, then the value of

(\sec x+{\cosec} x)^{2}

Solution:

Given,

\sin x-\cos x=0

⇒

\sin x=\cos x

⇒

{\sin x}/{\cos x}=1

∴

\tan x=1

⇒

x=45°

[\∵0°< x < 90°]

Now, the value of

(\sec x+{\cosec} x)^{2}

=(\sec 45°+{\cosec} 45°)^{2}

=(\√{2}+\√{2})^{2}=(2 \√{2})^{2}

=4 ×2=8

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