Math Olympiad Test: Trigonometry- 4 - Free MCQ with solutions Class 10

X sin³θ + Y cos³θ = sinθ cosθ ... (i)
X sinθ = Y cosθ ... (ii)
Using (ii) in (i), we get
⇒ Y cosqsin²θ + Y cos³θ = sinθcosθ
⇒ Y sin²θ + Y cos²θ = sinθ ⇒ Y = sinθ
∴ X sinθ = sinθ × cosθ ⇒ X = cosθ
∴ X² + Y² = 1

 ... (i)
 ... (ii)
Squaring and adding (i) and (ii), we get
 

sinx + sin²x = 1 (Given)
⇒ sinx = 1 – sin² x ⇒ sinx = cos² x
Now, cos⁸x + 2 cos⁶x + cos⁴x = sin⁴x + 2 sin³x + sin²x
= (sin²x + sinx)² = 1 [∵ (sinx + sin²x) = 1]

We have, l²m² (l² + m² + 3)
= (cosecθ – sinθ)² (secθ – cosθ)² {(cosecθ – sinθ)² + (secθ – cosθ)² + 3}



= cos⁶θ + sin⁶θ + 3 cos²θsin²θ
= {(cos²θ)3 + (sin²θ)3 + 3 cos²θsin²θ
= {(cos²θ + sin²θ)3 – 3 cos²θ sin²θ (cos²θ + sin²θ)} + 3 sin²θcos²θ
= {1 – 3 cos² sin²θ} + 3 cos²θ sin²θ = 1

Consider ΔABC in which ∠C = 90° and tan A = 1/√3.
Let BC = x.
Then, AC = √3x
By Pythagoras' theorem, we have,
AB² = AC² + BC² = (√3x)² + x² = 2x

 and sin B = 
∴ (sinA cosB + cosA sinB) 

(a) cosθ sinθ –  
= cosθ sinθ – sin³θ cosθ – cos³θ sinθ
= cosθ sinθ – cosθ sinθ (sin²θ + cos²θ)
= cosθ sinθ – cosθ sinθ = 0
(b) A and B are complementary angles
⇒ A + B = 90° ⇒ A = 90° – B
Now, taking R.H.S. we get 


= cosB = cos (90° – A) = sinA

(i) We have, x = a cos³θ and y = b sin³θ
∴ 

Hence,  cos²θ + sin²θ = 1
∴ P = 1.
(ii) We have, x = a secθcosφ
y = b secθsinφ and z = c tanθ

Hence,
(secθ cosφ)² + (secθ sinφ)² – (tanθ)²
= sec²θ – tan²θ = 1 + tan²θ – tan²θ = 1
∴ Q = 1.
(iii) cos A + cos² A = 1 (Given) ...(i)
∴ cos A = 1 – cos² A = sin² A
∴ sin² A + sin⁴ A = cos A + cos² A = 1
∴ R = 1.

(i) We know, 3(sinθ – cosθ)⁴ = 3((sinθ – cosθ)²)²
= 3(1² + 4sin²θcos²θ – 4sinθcosθ) ...(i)
and, 6(sinθ + cosθ)² = 6 + 12sinθcosθ ...(ii)
Also, 4(sin⁶θ + cos⁶θ) = 4((sin²θ)³ + (cos²θ)³) = 4(1) – 12sin²θcos²θ ...(iii)
Adding (i), (ii) and (iii), we get
3 + 12sin²θcos²θ – 12 sinθcosθ + 6+ 12 sinθ cosθ + 4 – 12 sin²θcos²θ
= 3 + 6 + 4 = 13 and 13 is independent of q.
(ii) We have,
cosecθ – sinθ = 
⇒ 
Similarly, secθ – cosθ =