Question

3. The value of 2(sin²45°+ cot²30°) – 6(cos²45°— tan²30°) is

(a) 6 (b) 3 (c) 2 (d) 4​

Answer 1

Answer:

2(sin^245°+cot^230°)-6(cos^2 45°-tan^230°)

=2{(1/√2)^2+(√3)^2}-6{(1/√2)^2-(1/√3)^2}

=2(1/2+3)-6(1/2-1/3)

=2×7/3-6×1/6

=14/3-1

=(14-3)/3

=11/3