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SSC CGL 18 Aug 2021 Shift 3 Paper

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Question : 69 of 100

Marks: +1, -0

If

\sin(A+B)=1

\cos(A-B)=\frac{\sqrt{3}}{2}, A+B\leq 90^{\circ}

A > B

, then the value of

\frac{5\sin^2 B+4\tan^2 A}{2\sin B\cos A}

is

$16\frac{1}{2}$
18
20
$26\frac{1}{2}$

Solution: 👈: Video Solution

Given,

\sin(A+B)=1

and

\cos(A-B)=\frac{\sqrt{3}}{2}

Then,

\sin(A+B)=\sin 90^{\circ}

\Rightarrow A+B=90^{\circ}

. . . (i)

\text{ and }\cos(A-B)=\cos 30^{\circ}

\Rightarrow A-B=30^{\circ}

. . . (ii)

From Eqs. (i) and (ii)

2A=120^{\circ}\Rightarrow A=60^{\circ}

and

B=30^{\circ}

\therefore\frac{5\sin^2 B+4\tan^2 A}{2\sin B\cos A}

=\frac{5\times\sin^2 30^{\circ}+4\tan^2 60^{\circ}}{2\sin 30^{\circ}\cdot\cos 60^{\circ}}

=\frac{5\times\left(\frac{1}{2}\right)^2+4\times(\sqrt{3})^2}{2\times\left(\frac{1}{2}\right)\cdot\frac{1}{2}}

=\frac{5}{4}+4\times 3\frac{15+48}{1\times\frac{1}{2}}=\frac{4}{\frac{1}{2}}

=\frac{53}{4}\times 2=\frac{53}{2}=26\frac{1}{2}.

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