RS Aggarwal Test: Introduction to Trigonometry - Grade 10 MCQ

we know

tan 30= 1/√3
cot60= 1/√3

tan30/cot60= 1/√3 ×√3/1= 1

Δ ABC is right angled at C, A + B + C = 180°
∴ A + B = 180° - 90° = 90° (∵ ∠C = 90°)
cos (A + B) = cos 90° = 0

cos (a + β) = 0 = cos 90°
⇒ α + β = 90°
sin (α - β) = sin (α - β + β - β)
= sin (α + β - 2β)
= sin (90° - 2β) = cos 2β

= 2
= R.H.S.

sin (45° + θ) - cos (45° - θ)
= sin {90° - (45° - θ)} - cos (45° - θ)
= cos (45° - θ) - cos (45° - θ) = 0

sin A = 1/2
so A = 30^ο 
cot A = cot30^ο = √3 

cosec A - cot A = 4/5  ........(i)
Also cosec² A - cot² A = 1
⇒ (cosec A - cot A) (cosec A - cot A) = 1

⇒ cosec A + cot A = 4/5   ......(ii)
From (i) and (ii), cosec A = 41/40

tan 1 . tan 2 . tan 3 ... tan 87 . tan 88 . tan 89 = LHS 
tan 1 . tan 2 . tan 3 ... tan (90 - 3 ) . tan ( 90 - 2 ) . tan ( 90 - 1)
tan 1 . tan 2  . tan 3 ... cot 3 . cot 2 . cot 1 

tan 1 . cot 1 . tan 2 . cot 2 . tan 3 . cot 3 ... tan 89 . cot 89 
1 x 1 x 1 x 1 x ... x 1 
As 1ⁿ = 1  = RHS 

sin x + cosec x = 2

We know that sin 60 =√3/2 and cos 30 = √3/2.
Therefore , Sin 60/cos 30= (√3/2)/(√3/2) = 1