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Exercise 3.3 · Q7

Q.Prove that tan⁡(π4+x)tan⁡(π4−x)=(1+tan⁡x1−tan⁡x)2\dfrac{\tan\left(\frac{\pi}{4}+x\right)}{\tan\left(\frac{\pi}{4}-x\right)} = \left(\dfrac{1+\tan x}{1-\tan x}\right)^2.

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Use the tangent addition formula on both angles in the left-hand side, simplify each fraction, then square the result to match the right-hand side. The identity follows from tan⁡(π4)=1\tan\left(\frac{\pi}{4}\right) = 1.

The heart of this proof lies in recognizing that π4\frac{\pi}{4} is special: its tangent equals 11. When we add or subtract xx from π4\frac{\pi}{4}, the tangent addition formula produces expressions that simplify beautifully because of this property. The strategy is to expand both the numerator and denominator of the left-hand side using the tangent sum and difference formulas, then observe the algebraic structure that emerges.

tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}

Proof:

  1. Expand the numerator using the tangent addition formula.

    We have tan⁡(π4+x)=tan⁡π4+tan⁡x1−tan⁡π4⋅tan⁡x\tan\left(\frac{\pi}{4} + x\right) = \frac{\tan\frac{\pi}{4} + \tan x}{1 - \tan\frac{\pi}{4} \cdot \tan x}.

    Since tan⁡π4=1\tan\frac{\pi}{4} = 1, this becomes:

tan⁡(π4+x)=1+tan⁡x1−tan⁡x\tan\left(\frac{\pi}{4} + x\right) = \frac{1 + \tan x}{1 - \tan x}

  1. Expand the denominator using the tangent difference formula.

    Similarly, tan⁡(π4−x)=tan⁡π4−tan⁡x1+tan⁡π4⋅tan⁡x\tan\left(\frac{\pi}{4} - x\right) = \frac{\tan\frac{\pi}{4} - \tan x}{1 + \tan\frac{\pi}{4} \cdot \tan x}.

    Again substituting tan⁡π4=1\tan\frac{\pi}{4} = 1:

tan⁡(π4−x)=1−tan⁡x1+tan⁡x\tan\left(\frac{\pi}{4} - x\right) = \frac{1 - \tan x}{1 + \tan x}

  1. Form the quotient of these two expressions.

    The left-hand side becomes:

tan⁡(π4+x)tan⁡(π4−x)=1+tan⁡x1−tan⁡x1−tan⁡x1+tan⁡x\frac{\tan\left(\frac{\pi}{4}+x\right)}{\tan\left(\frac{\pi}{4}-x\right)} = \frac{\frac{1 + \tan x}{1 - \tan x}}{\frac{1 - \tan x}{1 + \tan x}}

  1. Simplify by multiplying by the reciprocal. …

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