The value of {\cos 29^\circ {\cosec} 61^\circ \tan 45^\circ +2 \sin 35^\circ \sec 55^\circ}/{ 3 \sin ^{2} 42^\circ+3 \sin ^{2} 48^\circ}

SSC CHSL 19 Mar 2020 Shift 3 Paper

Question : 57 of 100

Marks: +1, -0

The value of

{\cos 29° {\cosec} 61° \tan 45° +2 \sin 35° \sec 55°}/{ 3 \sin ^{2} 42°+3 \sin ^{2} 48°}

Solution:

Given,

{\cos 29° {\cosec} 61° \tan 45° +2 \sin 35° \sec 55°}/{ 3 \sin ^{2} 42°+3 \sin ^{2} 48°}

={\cos 29° \×{1}/{\sin 61°} \×1+2 \sin 35° \×{1}/{\cos 55°}}/{3 \sin ^{2} 42°+3[\sin (90°-42°)]^{2}}

={\cos 29° \×{1}/{\cos 29°} \×1+2 \sin 35° \×{1}/{\sin 35°}}/{3 \sin ^{2} 42°+3 \cos ^{2} 42°}

[\∵\sin (90-\θ)=\cos \θ

and

\cos (90-\θ)=\sin \θ]

={1+2}/{3 \×(\sin ^{2} 42°+\cos ^{2} 42°)}

={3}/{3 \×1}

[\∵\sin ^{2} \θ+\cos ^{2} \θ=1]

= 1

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