Question

Find general solution of (1 – tan theta) (1 + tan 2theta )=1+ tan theta

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Answer 1

A solution of the equation (1−tanθ)(1+tanθ)sec

2

θ+2

tan

2

θ

=0 where θ lies in the interval (−π/2,π/2) is given by

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ANSWER

(1−tanθ)(1+tanθ).sec

2

θ+2

tan

2

θ

=0

(1−tan

2

θ).(1+tan

2

θ)+2

tan

2

θ

=0

(1−tan

4

θ)+2

tan

2

θ

=0

Let tan

2

θ=t.

Hence

1−t

2

+2

t

=0

t

2

−1=2

t

.

From inspection of graph we get

t

2

=9 is one solution.

t=3 since tan

2

θ=−3 is not possible.

tan

2

θ=3

tanθ=±

3

.

θ=±

3

π

.

Answer 2

Answer:

1

let theta=∆

[1+tan∆][1+2tan∆]

[1+tan∆][(1+2tan∆)÷1-tan^2∆]

【1-tan^2 ∆ =(1+tan2∆)(1-tan2∆)】

1+tan2∆ cancelled

there fore it becomes (1+tan2∆)÷(1-tan2∆)=1+tan∆

therefore

PROOVED♂™

NOTE:TAN2∆=2TAN∆÷1-TAN^2 ∆

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