SSC CGL 18 Aug 2021 Shift 3 Paper

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Question : 69 of 100

Marks: +1, -0

If

\sin (A+B)=1

and

\cos (A-B)=\;{√{3}}/{2}, A+B ≤ 90^{∘}

and

A > B

, then the value of

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

is

$16 \;{1}/{2}$
18
20
$26 \;{1}/{2}$

Solution:

Given,

\sin (A+B)=1

and

\cos (A-B)=\;{√{3}}/{2}

Then,

\;\; \sin (A+B)=\sin 90^{∘}

⇒\;A+B=90^{∘}

. . . (i)

\;\text " and "\;\;\cos (A-B)=\cos 30^{∘}

⇒\;A-B=30^{∘}

. . . (ii)
From Eqs. (i) and (ii)

2 A=120^{∘} ⇒ A=60^{∘}

and

B=30^{∘}

∴ \;{5 \sin ^{2} B+4 \tan ^{2} A}/{2 \sin B \cos A}

= \;\; \;{5 × \sin ^{2} 30^{∘}+4 \tan ^{2} 60^{∘}}/{2 \sin 30^{∘} ⋅ \cos 60^{∘}}

= \;\; \;{5 ×({1}/{2})^{2}+4 ×(√{3})^{2}}/{2 ×(\;{1}/{2}) ⋅ \;{1}/{2}}

= \;\; \;{5}/{4}+4 × 3 \;{15+48}/{1 × \;{1}/{2}}=\;{4}/{{1}/{2}}

= \;\; \;{53}/{4} × 2=\;{53}/{2}=26 \;{1}/{2} .

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