In a △ ABC , right angled at B if tan C= sqrt3 , then find sin²C+cos²Cover1+cot²C .

In a △ ABC , right angled at B if tan C= sqrt3 , then find…

In a $\triangle ABC$ , right angled at B if tan C= $\sqrt3$ , then find $sin^2C+cos^2C\over1+cot^2C$ .

Show answer & explanation

Correct answer: A

Explanation

The correct answer is option 1 i.e. $\frac{3}{4}$ .

tan C = √3 = tan 60°

⇒ C = 60°

Now, (sin²C + cos²C)/(1 + cot²C)

= 1/cosec²C [∵ sin²x + cos²x = 1 and 1 + cot²x = cosec²x]

= sin²C

= sin²60°

= (√3/2)²

= 3/4

Therefore, the value of the given expression is 3/4.