If 7 \cos ^{2} \theta+5 \sin ^{2} \theta-6=0, (0^{\circ} < \theta < 90^{\circ}), then what is the value of \sqrt{\frac{cosec \theta+\tan \theta}{\sec \theta-\cot \theta}}

SSC CGL 17 Aug 2021 Shift 3 Paper

Question : 66 of 100

Marks: +1, -0

7 \cos ^{2} θ+5 \sin ^{2} θ-6=0

(0^{∘} < θ < 90^{∘})

, then what is the value of

√{\;{cosec θ+\tan θ}/{\sec θ-\cot θ}}

Solution:

Given,

7 \cos ^{2} θ+5 \sin ^{2} θ-6=0

⇒ 5 \sin ^{2} θ+2 \cos ^{2} θ+5 \cos ^{2} θ-6=0

[ ∵ \sin ^{2} θ+\cos ^{2} θ=1]

⇒ 5+2 \cos ^{2} θ-6=0

⇒ \;\; \cos ^{2} θ=\;{1}/{2}

⇒ \;\; \cos θ=\;{1}/{√{2}}

\cos θ=\cos 45^{∘} ⇒ θ=45^{∘}

∴ √{\;{cosec θ+\tan θ}/{\sec θ-\cot θ}}

=√{\;{cosec 45^{∘}+\tan 45^{∘}}/{\sec 45^{∘}-\cot 45^{∘}}}=√{\;{√{2}+1}/{√{2}-1}}

Divide and Multiply by

√{2}+1

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

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