The value of \(\rm \frac{3(cosec^2 \,26^\circ-tan^2\,64^\circ)+(cot^2\,42^\circ-sec^2\,48^\circ)}{cot(22^\circ-\theta)-cosec^2(62^\circ+\theta)-tan(\theta+68^\circ)+tan^2(28^\circ-\theta)}\) is:

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The value of \(\rm \frac{3(cosec^2 \,26^\circ-tan^2\,64^\circ)+(cot^2\,42^\circ-sec^2\,48^\circ)}{cot(22^\circ-\theta)-cosec^2(62^\circ+\theta)-tan(\theta+68^\circ)+tan^2(28^\circ-\theta)}\) is:

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SSC CGL Tier 2 Quant Previous Paper 2 (Held On: 3 Feb 2022)

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3
4
–1
–2

Answer 1

Given:

\(\rm \frac{3(cosec^2 \,26^\circ-tan^2\,64^\circ)+(cot^2\,42^\circ-sec^2\,48^\circ)}{cot(22^\circ-θ)-cosec^2(62^\circ+θ)-tan(θ+68^\circ)+tan^2(28^\circ-θ)}\)

Formula used:

1 + tan²θ = sec²θ

⇒ tan²θ – sec²θ = -1

1 + cot²θ = cosec²θ

⇒ cot²θ – cosec²θ = -1

Calculation:

\(\rm \frac{3(cosec^2 \,26^\circ-tan^2\,64^\circ)+(cot^2\,42^\circ-sec^2\,48^\circ)}{cot(22^\circ-θ)-cosec^2(62^\circ+θ)-tan(θ+68^\circ)+tan^2(28^\circ-θ)}\)

Put θ = 0° as there is no given condition for θ

⇒ \(\rm \frac{3(sec^2 \,64^\circ-tan^2\,64^\circ)+(tan^2\,48^\circ-sec^2\,48^\circ)}{tan68^\circ-cosec^262^\circ-tan68^\circ+cot^262^\circ}\)

⇒ \(\rm \frac{3(1)+(-1)}{-cosec^262^\circ+cot^262^\circ}\)

⇒ \(\rm \frac{3-1}{-1}\)

⇒ \(\rm \frac{2}{-1}\)

⇒ –2

∴ The required value is –2.

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The value of \(\rm \frac{3(cosec^2 \,26^\circ-tan^2\,64^\circ)+(cot^2\,42^\circ-sec^2\,48^\circ)}{cot(22^\circ-\theta)-cosec^2(62^\circ+\theta)-tan(\theta+68^\circ)+tan^2(28^\circ-\theta)}\) is:

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