Trigonometrical Equations 2 Question 4

4. Find all values of θ in the interval π2,π2 satisfying the equation (1tanθ)(1+tanθ)sec2θ+2tan2θ=0.

(1996, 2M)

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Answer:

Correct Answer: 4. θ=±π/3

Solution:

  1. Given, (1tanθ)(1+tanθ)sec2θ+2tan2θ=0

(1tan2θ)(1+tan2θ)+2tan2θ=0

1tan4θ+2tan2θ=0

Put tan2θ=x

1x2+2x=0

x21=2x

NOTE 2x and x21 are uncompatible functions, therefore we have to consider range of both functions.

Curves y=x21 and y=2x

It is clear from the graph that two curves intersect at one point at x=3,y=8.

Therefore, tan2θ=3

tanθ=±3θ=±π3



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