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Question 154 of 175

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

(b) Evaluate: ∫xcos⁡x dx\displaystyle\int x \cos x\, dx
Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2022Subjective· 5mImportance★★★★★
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Reducing every ratio to a function of θ\theta using standard identities collapses the whole expression to cos⁡2θcot⁡θ\cos^2\theta\cot\theta.

Reduce each factor using standard identities:

cot⁡(180°+θ)=cot⁡θ\cot(180°+\theta) = \cot\theta (tangent, and hence cotangent, repeats with a sign-preserving shift of 180°180°).

sin⁡(90°−θ)=cos⁡θ\sin(90°-\theta) = \cos\theta.

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta (cosine is even).

sin⁡(270°+θ)=sin⁡270°cos⁡θ+cos⁡270°sin⁡θ=(−1)cos⁡θ+0=−cos⁡θ\sin(270°+\theta) = \sin270°\cos\theta+\cos270°\sin\theta = (-1)\cos\theta+0 = -\cos\theta.

tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan\theta (tangent is odd).

cosec(360°+θ)=cosec θ\text{cosec}(360°+\theta) = \text{cosec}\,\theta (period 360°360°).

Numerator =cot⁡θ⋅cos⁡θ⋅cos⁡θ=cot⁡θcos⁡2θ= \cot\theta\cdot\cos\theta\cdot\cos\theta = \cot\theta\cos^2\theta.

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