Question

eve that:
(3 - 4sin^2A).(1 - 3tan^2 A) = (3-tan^2 A).(4cos^2 A-3)​

Answer 1

Answer:

L.H.s. (3-4sin^2A ). ( 1-3tan^2A)

= 3-9tan^2A -4sin^2A+12 sin^2A.tan^2A

= 3tan^2A (4sin^2A-3) + 1(3-4sin^2A)

= (4sin^2A-3) (3tan^2A -1)

R.H.s.

(3-tan^2A) (4cos^2A-3)

12cos^2A-9 -4sin^2A+3tan^2A

12-12sin^2A-9 -4sin^2A+3tan^2A

3-12 sin^2A + 3 tan^2A -4sin^2A

-1(4sin^2A-3) -3 (4sin^2A-tan^2A)

Answer 2

given

(3-4sin²A)(1-3tan²A)=(3-tan²A)(4cos²A-3)

➡(3-4sin²A)/(4cos²A-3)=(3-tan²A)/(1-3tan²A)

LHS

$\frac{3 - 4 {sin}^{2} a}{4 {cos}^{2}a - 3}$

Now

$\frac{3sina - 4 {sin}^{3} a}{4 {cos}^{3} a - 3cosa} \times \frac{cosa}{sina} \\ \\ = tan3a \times cota \\ \\ = \frac{3tana - {tan}^{3} a}{1 - 3 {tan}^{2}a } \times cota \\ \\ = \frac{3tanacota - {tan}^{3} acota}{1 - 3 {tan}^{2}a }$

$= \frac{3 - {tan}^{2}a }{1 - 3{tan}^{2}a }$

RHS