Question

32. The value of (sec?e + 2 tan cote-tan? e) is RRB NTPC (Phase 1) 2016 (1) 0 (2) 1 (3) 2 (4) 3

11th - 12th Class

Maths

Solution

106

4.0 (1 ratings)

32. (4) By trigonometric identities [ begin{aligned} 1+tan ^{2} theta &=sec ^{2} theta sec ^{2} theta+2 tan theta cot theta-tan ^{2} theta &=1+2 tan theta cot theta end{aligned} ] By reciprocal relation of Trigonometric ratio ( tan theta=frac{1}{cot theta} Rightarrow 1+2 times 1=1+2=3 )