Question

7.If tan A and tanB are the roots of the quadraticequation, 3x^2-10x-25=0, then the value of3 sin^2 (A+B)-10 sin(A+B).COS(A+B)-25 cos^2(A+B)is :(JEE MAIN (Online) - 2018](A) 10 (B) -10 (C) 25 (0) --25​

Answer 1

Solution :

The given equation is

3x² - 10x - 25 = 0

or, 3x² - 15x + 5x - 25 = 0

or, 3x (x - 5) + 5 (x - 5) = 0

or, (x - 5) (3x + 5) = 0

Either x - 5 = 0 or, 3x + 5 = 0

i.e., x = 5, - 5/3

Let us consider:

tanA = 5 and tanB = - 5/3 [ ATQ ]

Now, tan(A + B)

= (tanA + tanB)/(1 - tanA tanB)

= {5 + (- 5/3)}/{1 - 5 (- 5/3)}

= (5 - 5/3)/(1 + 25/3)

= (15 - 5)/(3 + 25)

= 10/28

= 5/14

Thus, hypotenuse = √(5² + 14²) = √221

Then, sin(A + B) = 5/√221

and cos(A + B) = 14/√221

Now, 3 sin²(A + B) - 10 sin(A + B) cos(A + B) - 25 cos²(A + B)

= 3 (5/√221)² - 10 (5/√221) (14/√221) - 25 (14/√221)²

= 3 (25/221) - 10 (70/221) - 25 (196/221)

= 75/221 - 700/221 - 4900/221

= - 5525/221

= - 25

Therefore, option (D) is correct.

Answer 2

Answer:

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