Previous Year NEET Question- Trigonometric Functions

• 2015:

In a right triangle ABC, right angled at B, we have:

sin A = 1/√3

Since, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, we have:

a^2 + b^2 = c^2

Substituting the value of sin A, we get:

b^2 = c^2 - a^2 = c^2 - (1/√3)^2 = ( (√3)^2 - (1/√3)^2 ) = 3 - 1/3 = 8/3

Therefore, cos C = b/c = √3/3.

2016:

We have:

sin A + sin B = √3/2 (equation valid for specific angles A and B)
cos A + cos B = √2/2

Adding the two equations, we get:

2 sin (A + B)/2 * cos (A - B)/2 = (√3 + 1)/2

Dividing both sides