6 cos θ - 7 sin θ = 0 ⇒ (7cos2\theta+6sin2\theta)^2

TS ICET 23 May 2019 Shift 1 Paper

Question : 120 of 200

Marks: +1, -0

6 cos θ - 7 sin θ = 0 ⇒

(7cos2θ+6sin2θ)^2

Solution:

6 \cos θ-7 \sin 7 θ=0

6 \cos θ=7 \sin θ ⇒ \;{\sin θ}/{\cos θ}=\;{6}/{7} ⇒ \tan θ=\;{6}/{7}

(7 \cos 2 θ+6 \sin 2 θ)=7 \;{(1-\tan ^{2} θ)}/{1+\tan ^{2} θ}+6 \;{(2 \tan θ)}/{1+\tan ^{2} θ}

=\;{7(1-{36}/{49})+12 {6}/{7}}/{1+\;{36}/{7}}=\;{13+72}/{85} × \;{49}/{7}=7

∴ Required answer

= 7^2 = 49

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