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NCERT Exemplar · Q44

Q.The value of cot⁡(π4+θ)cot⁡(π4−θ)\cot\left(\dfrac{\pi}{4} + \theta\right)\cot\left(\dfrac{\pi}{4} - \theta\right) is
(A) −1-1
(B) 00
(C) 11
(D) Not defined

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Using the cotangent addition and subtraction formulas, the product simplifies to 11 because the numerator and denominator cancel identically. The final value is 11.

The key to this problem lies in how the cotangent function behaves when you add or subtract angles. Unlike sine or cosine, cotangent has a reciprocal relationship with tangent, and the formulas for cot⁡(A±B)\cot(A \pm B) are less commonly memorised. But that’s fine — we can derive them from the tangent formulas.

When you see π4+θ\frac{\pi}{4} + \theta and π4−θ\frac{\pi}{4} - \theta, notice that these two angles are complementary in a shifted sense: their sum is π2\frac{\pi}{2}. That’s a strong hint that the product might simplify to something neat. Let’s work it out.

  1. Write each cotangent in terms of tangent. Since cot⁡x=1tan⁡x\cot x = \frac{1}{\tan x}, we have:

cot⁡(π4+θ)=1tan⁡(π4+θ),cot⁡(π4−θ)=1tan⁡(π4−θ).\cot\left(\frac{\pi}{4} + \theta\right) = \frac{1}{\tan\left(\frac{\pi}{4} + \theta\right)}, \quad \cot\left(\frac{\pi}{4} - \theta\right) = \frac{1}{\tan\left(\frac{\pi}{4} - \theta\right)}.

So the product becomes:

1tan⁡(π4+θ)⋅tan⁡(π4−θ).\frac{1}{\tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{\pi}{4} - \theta\right)}.

  1. Apply the tangent addition and subtraction formulas. Recall:

tan⁡(A+B)=tan⁡A+tan⁡B1−tan⁡Atan⁡B,tan⁡(A−B)=tan⁡A−tan⁡B1+tan⁡Atan⁡B.\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}, \quad \tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}.

Here A=π4A = \frac{\pi}{4} and B=θB = \theta. Since tan⁡π4=1\tan\frac{\pi}{4} = 1, we get:

tan⁡(π4+θ)=1+tan⁡θ1−tan⁡θ,\tan\left(\frac{\pi}{4} + \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta},

tan⁡(π4−θ)=1−tan⁡θ1+tan⁡θ.\tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 - \tan\theta}{1 + \tan\theta}.

  1. Multiply the two tangents.

tan⁡(π4+θ)⋅tan⁡(π4−θ)=1+tan⁡θ1−tan⁡θ⋅1−tan⁡θ1+tan⁡θ.\tan\left(\frac{\pi}{4} + \theta\right) \cdot \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1 + \tan\theta}{1 - \tan\theta} \cdot \frac{1 - \tan\theta}{1 + \tan\theta}.

The numerator and denominator are identical products — they cancel completely:

=1.= 1.

  1. Now take the reciprocal to get the cotangent product. Since the product of the tangents is 11, the product of the cotangents is:

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