Question

Question Number 38
Please tell the answer

Answer 1

(sec 39/cosec 51) +(2/√3*tan 17*tan 38*tan 52*tan 73)- 3(sin²31 + sin²59)

cosec(90-39)/cosec 51

= cosec 51/cosec 51 = 1.........(1)

2/√3* cot(90-17) * cot( 90-38) * tan 52* tan 73

2/√3 * cot 73 * tan 73 * tan 52* cot 52

2/√3 * 1 .........( because cotθ = 1/tanθ)

= 2/√3......(2)

-3(sin²31 + sin²59)

-3( cos² (90-31) + sin²59)

-3(1) .....(because sin²θ + cos²θ = 1)

= -3.....(3)

Adding (1) + (2) + (3)

=1 + 2/√3 -3

=1 + 2√3/3 -3

=(3 + 2√3-9)/3

=(-6 + 2√3)/3

= - 2 + 2√3/3