Question: If exp {(tan2 x – tan4 x + tan6 x – tan8 x + ….) log<sub...

Question

If exp {(tan² x – tan⁴ x + tan⁶ x – tan⁸ x + ….) log_e 16}, 0 < x <p/4, satisfies the quadratic equation x² – 3x + 2 = 0, then value of cos² x + cos⁴x is-

Answer 1

Solution

We have

tan² x – tan⁴ x + tan⁶ x – tan⁸ x + …….

= $\frac { \tan ^ { 2 } x } { 1 - \left( - \tan ^ { 2 } x \right) }$ = $\frac { \tan ^ { 2 } x } { \sec ^ { 2 } x }$ = sin² x

\ y = exp {(tan² x – tan⁴ x + tan⁶ x – tan⁸x + ….) log_e 16}

= exp {(sin² x)log_e16} = exp {log_e }

=

As y satisfies x³ – 3x + 2 = 0, we get

y = 1 or y = 2.

̃ = 1 or = 2

Since 0 < x < p/4, 0 < sin x < 1/ $\sqrt { 2 }$

̃ 0 < sin² x < ½

̃ = 1 is not possible.

Thus, = 2

̃ sin² x = ¼

Thus, cos² x + cos⁴ x = (1 – sin² x) + (1 – sin² x)²= 21/16.

Question: If exp {(tan<sup>2</sup> x – tan<sup>4</sup> x + tan<sup>6</sup> x – tan<sup>8</sup> x + ….) log<sub...

Solution