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Question 170 of 175

Q.(a) Prove that cot⁡(180∘+θ)sin⁡(90∘−θ)cos⁡(−θ)sin⁡(270∘+θ)tan⁡(−θ)cosec(360∘+θ)=cos⁡2θcot⁡θ\dfrac{\cot(180^\circ+\theta)\sin(90^\circ-\theta)\cos(-\theta)}{\sin(270^\circ+\theta)\tan(-\theta)\text{cosec}(360^\circ+\theta)} = \cos^2\theta\cot\theta OR

(b) If y=sin⁡−1x1−x2y = \dfrac{\sin^{-1}x}{\sqrt{1-x^2}}, show that (1−x2)y2−3xy1−y=0(1-x^2)y_2 - 3xy_1 - y = 0
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2025Subjective· 5mImportance★★★★★
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Reduce each allied angle to a function of θ using the standard sign rules, then simplify — the expression collapses to cos²θ cotθ.

Use the allied-angle (reduction) formulas:

  • cot⁡(180∘+θ)=cot⁡θ\cot(180^\circ+\theta) = \cot\theta (180°+θ lies in the 3rd quadrant, where tan/cot are positive)
  • sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ-\theta) = \cos\theta
  • cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta (cosine is even)
  • sin⁡(270∘+θ)=−cos⁡θ\sin(270^\circ+\theta) = -\cos\theta (270°+θ lies in the 4th quadrant, where sine is negative, and 270° flips sine and cosine)
  • tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta (tangent is odd)
  • cosec(360∘+θ)=cosec θ\text{cosec}(360^\circ+\theta) = \text{cosec}\,\theta (360° is a full period)

Numerator:

cot⁡θ⋅cos⁡θ⋅cos⁡θ=cos⁡2θcot⁡θ\cot\theta\cdot\cos\theta\cdot\cos\theta = \cos^2\theta\cot\theta

Denominator: …

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