Question

4 cos A - 3 sec A = 2 tan A
then A=?​

Answer 1

Answer:

4 cos Ø - 3 sec Ø = 2 tan Ø

=> 4 cos Ø - 3/cos Ø = 2 sin Ø/cos Ø [°.° sec Ø = 1/cos Ø]

=> 4 cos^2 Ø - 3 = 2 sin Ø/ cos Ø × cos Ø

=> 4 cos^2 Ø - 3 = 2 sin Ø

=> 4 ( 1 - sin^2 Ø ) - 3 - 2 sin Ø = 0 [°.° cos^2 Ø = 1 - sin^2 Ø]

=> 4 - 4 sin^2 Ø - 3 - 2 sin Ø = 0

=> 4 sin^2 Ø + 2 sin Ø - 1 = 0

Now, treat sin^2 Ø as a variable such as x

=> 4x^2 + 2x - 1 = 0

$\tt{x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}}$

=> $\tt{x = \frac{-2 \pm \sqrt{4 + 4}}{8}}$

=> $\tt{x = \frac{-1 \pm 2\sqrt{2}}{4}}$

There is no standard value of Ø which satisfies the value of x, or sin Ø.

_________________

Answer 2

Answer:

happy birthday

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