Question

IF 3SINX + 4COSX= 5
THEN FIND 6.TANX/2 - 9.TAN^2X/2=???
IF ANYONE SOLVES THIS IN TWO METHODS I WILL MARK ALL THEIR ANS AS BRAINLYIST AND I'LL FOLLOW THEM☺️​​

Answer 1

It has given that, 3sinx + 4cosx = 5

we have to find the 6 tan²(x/2) - 9 tan²(x/2)

solution : method 1 :

3sinx + 4cosx = 5

using sinx = 2tan(x/2)/(1 + tan²(x/2))

cosx = (1 - tan²(x/2))/(1 + tan²(x/2))

now 3[2tan(x/2)/(1 + tan²(x/2))] + 4[(1 - tan²(x/2))/(1 + tan²(x/2))] = 5

⇒6tan(x/2)/{1 + tan²(x/2)} + {4 - 4tan²(x/2)}/{1 + tan²(x/2)} = 5

⇒6tan(x/2) + 4 - 4tan²(x/2) = 5 + 5 tan²(x/2)

⇒6 tan(x/2) - 9tan²(x/2) = 5 - 4 = 1

⇒6 tan(x/2) - 9 tan²(x/2) = 1 [ Ans]

method 2 : given, 3sinx + 4cosx = 5

let sinx = 3/5 and cosx = 4/5

then, 3 × 3/5 + 4 × 4/5 = 5 is satisfied ..

now sinx/cosx = (3/5)/(4/5)

⇒tanx = 3/4

using formula, tanx = 2tan(x/2)/(1 - tan²(x/2))

so, 2tan(x/2)/(1 - tan²(x/2)) = 3/4

⇒8tan(x/2) = 3 - 3tan²(x/2)

⇒3tan²(x/2) + 8 tan(x/2) - 3 = 0

⇒tan(x/2) = {-8 ± √(64 + 36)}/6 = -4, 1/3

taking tan(x/2) = 1/3

so, 6tan(x/2) - 9tan²(x/2) = 6 × 1/3 - 9 × 1/9

= 2 - 1 = 1 [ Ans ]

So the answer is 6 tan²(x/2) - 9 tan²(x/2) = 1

Answer 2

Given:-

• 3Sin X + 4CosX= 5.

To find:-

• Value of Tan X/2 - 9Tan^2X/2.

Solution:-

Formula :

• Sin x = 2 tan ( x/2) / 1+ tan² (x/2 )

• cos x = 1 - tan ² (x/2)  / 1 + tan²(x/2)

Given , 3sinx + 4cosx = 5

$=> 3\frac{2tanx/2}{1+tan^2 x/2} + 4 \frac{1-tan^2 x/2 }{1+ tan^2 x/2} =5 \\\\\\\\=> \frac{6\: tan\: x/2 \: \: + \: 4 - 4 \: tan^2 \: x/2}{1+tan^2 \: x/2} =5\\\\\\=> 6\:tan\: \frac{x}{2} + 4 - 4\: tan^2\frac{x}{2} = 5 + 5tan^2\frac{x}{2} \\\\\\=> 6\: tan \: \frac{x}{2} + 4 - 4\: tan^2 \: \frac{x}{2} - 5- 5 \: tan^2 \frac{x}{2} = 0 \\\\=> 9\: tan^2 \frac{x}{2} - 6\: tan\frac{x}{2} +1 = 0$

=> (3+ tan x/2) ²  - 2×3 tan x/2 + 1 = 0

=> (3 tan x/2  - 1 )² = 0

=> tan x/2  =  1/3

Now,

=> 6 tan x/2  - 9 tan² x/2  = 6 ×1/3 -9(1/3)²

           =  2-1 = 1.

Hence, the required answer is 1.

_________________________________