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Exercise 3.3 · Q9

Q.Prove that cos⁡(3π2+x)cos⁡(2π+x)[cot⁡(3π2−x)+cot⁡(2π+x)]=1\cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi+x)\left[\cot\left(\frac{3\pi}{2}-x\right) + \cot(2\pi+x)\right] = 1.

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The expression simplifies to 11 by reducing each trigonometric function to its simplest form using periodicity and quadrant rules, then applying the identity cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos\theta}{\sin\theta} and cancelling common terms.

We need to prove:

cos⁡(3π2+x)cos⁡(2π+x)[cot⁡(3π2−x)+cot⁡(2π+x)]=1.\cos\left(\frac{3\pi}{2}+x\right)\cos(2\pi+x)\left[\cot\left(\frac{3\pi}{2}-x\right) + \cot(2\pi+x)\right] = 1.

The key is to simplify each term using standard reduction formulas. Trigonometric functions of angles like 3π2±x\frac{3\pi}{2} \pm x and 2π±x2\pi \pm x can be expressed in terms of functions of xx alone, using the unit circle or the ASTC rule (All, Sine, Tangent, Cosine). Let’s do this step by step.

  1. Simplify cos⁡(3π2+x)\cos\left(\frac{3\pi}{2}+x\right) 3π2\frac{3\pi}{2} is on the negative y-axis. Adding xx (a small positive angle) moves into the fourth quadrant, where cosine is positive. The reference angle is xx, and since cosine is sin⁡\sin shifted, we have:

cos⁡(3π2+x)=sin⁡x.\cos\left(\frac{3\pi}{2}+x\right) = \sin x.

(Check: at x=0x=0, cos⁡(3π/2)=0=sin⁡0\cos(3\pi/2)=0 = \sin 0; at x=π/2x=\pi/2, cos⁡(2π)=1=sin⁡(π/2)\cos(2\pi)=1 = \sin(\pi/2) — consistent.)

  1. Simplify cos⁡(2π+x)\cos(2\pi+x) 2π2\pi is a full rotation, so cosine repeats every 2π2\pi:

cos⁡(2π+x)=cos⁡x.\cos(2\pi+x) = \cos x.

  1. Simplify cot⁡(3π2−x)\cot\left(\frac{3\pi}{2}-x\right) 3π2−x\frac{3\pi}{2} - x: starting from 3π2\frac{3\pi}{2} (negative y-axis) and moving backward by xx puts us in the third quadrant (since 3π2−x\frac{3\pi}{2} - x is between π\pi and 3π2\frac{3\pi}{2} for small xx). In the third quadrant, cot⁡\cot (cos/sin) is positive because both sine and cosine are negative. The reference angle is xx, and:

cot⁡(3π2−x)=tan⁡x.\cot\left(\frac{3\pi}{2}-x\right) = \tan x.

(Why? Because cot⁡(3π2−θ)=tan⁡θ\cot(\frac{3\pi}{2} - \theta) = \tan\theta is a standard reduction: 3π2−θ\frac{3\pi}{2} - \theta gives the cofunction.)

  1. Simplify cot⁡(2π+x)\cot(2\pi+x) 2π2\pi is a full period for cotangent as well (since cot⁡\cot has period π\pi, but 2π2\pi is a multiple):

cot⁡(2π+x)=cot⁡x.\cot(2\pi+x) = \cot x.

Now substitute these into the original expression:

(sin⁡x)(cos⁡x)[tan⁡x+cot⁡x].(\sin x)(\cos x)\left[\tan x + \cot x\right].

  1. Simplify the bracket Write tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x} and cot⁡x=cos⁡xsin⁡x\cot x = \frac{\cos x}{\sin x}:

tan⁡x+cot⁡x=sin⁡xcos⁡x+cos⁡xsin⁡x=sin⁡2x+cos⁡2xsin⁡xcos⁡x=1sin⁡xcos⁡x.\tan x + \cot x = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x \cos x} = \frac{1}{\sin x \cos x}.

  1. Multiply everything …

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