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Exercise 3.3 · Q4

Q.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)\,\operatorname{cosec}(360^\circ+\theta)}=\cos^2\theta\cot\theta.

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Every factor here is an allied angle of θ\theta; substituting each one from the §3.4.3 table collapses the whole expression to cos⁡2θcot⁡θ\cos^2\theta\cot\theta.

Step 1. Reduce the numerator's three factors.

cot⁡(180∘+θ)=cot⁡θ\cot(180^\circ+\theta)=\cot\theta (an even multiple of 90∘90^\circ away — name unchanged; 180∘+θ180^\circ+\theta lies in Quadrant III, where tan/cot are positive, matching a ++ sign).

sin⁡(90∘−θ)=cos⁡θ\sin(90^\circ-\theta)=\cos\theta (odd multiple of 90∘90^\circ — co-changes to cosine; 90∘−θ90^\circ-\theta is acute, so Quadrant I, sign ++).

cos⁡(−θ)=cos⁡θ\cos(-\theta)=\cos\theta (negative-angle identity).

So the numerator is cot⁡θ⋅cos⁡θ⋅cos⁡θ=cot⁡θcos⁡2θ\cot\theta\cdot\cos\theta\cdot\cos\theta=\cot\theta\cos^2\theta.

Step 2. Reduce the denominator's three factors.

sin⁡(270∘+θ)=−cos⁡θ\sin(270^\circ+\theta)=-\cos\theta (odd multiple of 90∘90^\circ — co-changes to cosine; 270∘+θ270^\circ+\theta falls in Quadrant IV where sine is negative, sign −-).

tan⁡(−θ)=−tan⁡θ\tan(-\theta)=-\tan\theta (negative-angle identity).

cosec⁡(360∘+θ)=cosec⁡θ\operatorname{cosec}(360^\circ+\theta)=\operatorname{cosec}\theta (a full rotation added — value unchanged). …

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