The value of {\sec ^{2} 60^{\circ} \cos ^{2} 45^{\circ}+cosec^{2} 30^{\circ}}/{\cot 30^{\circ} \sec ^{2} 45^{\circ}-cosec^{2} 30^{\circ} \tan 45^{\circ}} is

SSC CGL 24 Aug 2021 Shift 2 Paper

Question : 61 of 100

Marks: +1, -0

The value of

\;{\sec ^{2} 60^{∘} \cos ^{2} 45^{∘}+cosec^{2} 30^{∘}}/{\cot 30^{∘} \sec ^{2} 45^{∘}-cosec^{2} 30^{∘} \tan 45^{∘}}

Solution:

\;{\sec ^{2} 60^{∘} \cos ^{2} 45^{∘}+cosec^{2} 30^{∘}}/{\cot 30^{∘} × \sec ^{2} 45^{∘}-cosec^{2} 30^{∘} \tan 45^{∘}}

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

[ ∵ \sec 60^{∘}=2, \cos 45^{∘}=\;{1}/{√{2}}

cosec 30^{∘}=2, \cot 30^{∘}=√{3}

\sec 45^{∘}=√{2}, \tan 45^{∘}=1]

=\;{4 × {1}/{2}+4}/{√{3} × 2-4}

=\;{2+4}/{2 √{3}-4}=\;{6}/{2(√{3}-2)}

=\;{3}/{(√{3}-2)}

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

=-3(2+√{3})

[ ∵ (a+b)(a-b)=a^{2}-b^{2}]

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