Math Olympiad Test: Trigonometry- 2 - Class 10 MCQ

We have 


∴ Both 1 and 2 are correct.

We know that, 1 radian = 180°/π = 57°30’ approx
57° lies between 0 and 90 degrees and since in first quadrant sin θ increases when θ increases
⇒ sin 1°< sin 1.

Here sin θ − cosθ = 3/5
Squaring both sides, we get
sin²θ + cos²θ - 2 sin θ cos θ = 9/25
⇒ 1 - 2 sin θ cos θ = 9/25
⇒ 2 sin θ cos θ = 1 - 9/25 = 16/25
⇒ sin θ cos θ = 8/25

Here sin θ cos(90 - θ) + cos θ · sin (90 - θ)
= sin θ · sin θ + cos θ · cos θ
= sin2θ + cos2θ = 1

Given α = sec θ - tan  θ (i)
and b = sec  θ + tan  θ (ii)
Multiplying (i) and (ii) we get
sec²θ - tan²θ = α - b
⇒ 1 = ab
⇒ α = 1/b

We have tan 15 tan 20 tan 70 tan 75
= tan 15 tan 20 tan (90 - 20) tan (90 - 15)
= tan 15 tan 20 - cot 20 cot 15

= 1

a⁴ − b⁴ = (a² − b²)(a² + b²)

Recall the range of cosθ.

We have,
x = a(cosecθ + cotθ)  ……. (1)
y = b(cosecθ − cotθ)  …….. (2)
From equation (1) and (2), we get
xy = ab(cosecθ + cotθ)(cosec − cotθ)
xy = ab(cosec2θ − cot2θ)
xy = ab[∵cosec2x − cot2x = 1]
Hence, this is the answer.

We have

Here log sin 0 + log sin 1 + log sin 2 + . . . + log sin 90
= log (sin 0 × sin 1 × sin 2 . . . sin 90)
= log (0) = 0

sin²20 + cos²160 - tan²45 = sin²20 + cos²160 - tan²45
= sin²(180 - 160) + cos²160 - tan²45
= sin²160 + cos²160 - tan²45
= 1 - 1 = 0