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:

This question was previously asked in
SSC CGL Tier 2 Quant Previous Paper 2 (Held On: 3 Feb 2022)
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  1. 3
  2. 4
  3. –1
  4. –2

Answer (Detailed Solution Below)

Option 4 : –2
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Detailed Solution

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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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