The value of {\tan 50^{\circ}+\sec 50^{\circ}}/{\cot 40^{\circ}+cosec 40^{\circ}}+\cos ^{2} 65^{\circ} +cosec 40^{\circ} + \sin 65^{\circ} \cos 25^{\circ}+\tan 30^{\circ} \text{ is }

SSC CHSL 5 Aug 2021 Shift 3 Paper

Section: Quantitative Aptitude

Question : 71 of 100

Marks: +1, -0

The value of

\;{\tan 50^{∘}+\sec 50^{∘}}/{\cot 40^{∘}+cosec 40^{∘}}+\cos ^{2} 65^{∘}

+cosec 40^{∘}

+ \sin 65^{∘} \cos 25^{∘}+\tan 30^{∘} \;\text " is "\;

Solution:

\;{\tan 50^{∘}+\sec 50^{∘}}/{\cot 40^{∘}+cosec 40^{∘}}+\cos ^{2} 65^{∘}

+\sin 65^{∘} × \cos 25^{∘}+\tan 30^{∘}

=\;{\tan (90^{∘}-40^{∘})+\sec (90^{∘}-40^{∘})}/{\cot 40^{∘}+cosec 40^{∘}}

+\cos ^{2} 65^{∘}+\sin 65^{∘}

× \cos (90^{∘}-65^{∘})+\tan 30^{∘}

=\;{\cot 40^{∘}+cosec 40^{∘}}/{\cot 40^{∘}+cosec 40^{∘}}

+\cos ^{2} 65^{∘}+\sin ^{2} 65^{∘}+\tan 30^{∘}

=1+1+\;{1}/{√{3}}

=2+\;{1}/{√{3}}=\;{2 √{3}+1}/{√{3}}

=\;{(2 √{3}+1)}/{√{3}} × \;{√{3}}/{√{3}}=\;{6+√{3}}/{3}

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