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NCERT Exemplar · Q42

Q.The value of tan⁡3A−tan⁡2A−tan⁡A\tan 3A - \tan 2A - \tan A is equal to
(A) tan⁡3A tan⁡2A tan⁡A\tan 3A\,\tan 2A\,\tan A
(B) −tan⁡3A tan⁡2A tan⁡A-\tan 3A\,\tan 2A\,\tan A
(C) tan⁡A tan⁡2A−tan⁡2A tan⁡3A−tan⁡3A tan⁡A\tan A\,\tan 2A - \tan 2A\,\tan 3A - \tan 3A\,\tan A
(D) None of these

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By using the angle relationship 3A=2A+A3A = 2A + A and applying the tangent addition formula, the expression tan⁡3A−tan⁡2A−tan⁡A\tan 3A - \tan 2A - \tan A simplifies to tan⁡3A tan⁡2A tan⁡A\tan 3A\,\tan 2A\,\tan A.

The problem asks us to simplify the expression tan⁡3A−tan⁡2A−tan⁡A\tan 3A - \tan 2A - \tan A. The key to solving this type of problem lies in recognizing the relationship between the angles involved. Here, we have angles AA, 2A2A, and 3A3A. Notice that 3A3A is the sum of 2A2A and AA. This immediately suggests using the tangent addition formula.

The tangent addition formula is:

tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡xtan⁡y\tan(x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y}

This formula is fundamental for relating the tangent of a sum of angles to the tangents of the individual angles. By setting x=2Ax=2A and y=Ay=A, we can express tan⁡(3A)\tan(3A) in terms of tan⁡(2A)\tan(2A) and tan⁡(A)\tan(A), which will allow us to manipulate the given expression.

Let's proceed with the steps:

  1. Establish the angle relationship: We start by noting the direct relationship between the angles:

3A=2A+A3A = 2A + A

  1. Apply the tangent function to both sides: Taking the tangent of both sides of this equation gives us:

tan⁡(3A)=tan⁡(2A+A)\tan(3A) = \tan(2A + A)

  1. Use the tangent addition formula: Now, we apply the tangent addition formula to the right-hand side, with x=2Ax = 2A and y=Ay = A:

tan⁡(3A)=tan⁡(2A)+tan⁡(A)1−tan⁡(2A)tan⁡(A)\tan(3A) = \frac{\tan(2A) + \tan(A)}{1 - \tan(2A)\tan(A)}

  1. Rearrange the equation: To simplify and get closer to the form tan⁡3A−tan⁡2A−tan⁡A\tan 3A - \tan 2A - \tan A, we multiply both sides of the equation by the denominator (1−tan⁡(2A)tan⁡(A))(1 - \tan(2A)\tan(A)):

tan⁡(3A)(1−tan⁡(2A)tan⁡(A))=tan⁡(2A)+tan⁡(A)\tan(3A) (1 - \tan(2A)\tan(A)) = \tan(2A) + \tan(A)

  1. Expand the left-hand side: Distribute tan⁡(3A)\tan(3A) across the terms in the parenthesis:

tan⁡(3A)−tan⁡(3A)tan⁡(2A)tan⁡(A)=tan⁡(2A)+tan⁡(A)\tan(3A) - \tan(3A)\tan(2A)\tan(A) = \tan(2A) + \tan(A)

  1. Isolate the desired expression: …

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