Applications of Trigonometric Identities - Free MCQ Practice Test

7sin²x+3cos² x=4
7sin²x+3(1-sin²x)=4
7sin²x+3-3sin²x=4
4sin²x=4-3
4sin²x=1
sin²x=¼
sinx=½
Cosec x=1/sinx=2
Cos x= 
Sec x= 1/cos x= 
Cosec x + sec x=2+ 

Answer: (c) √(b²-a²)/b

Explanation: cos X = a/b

By trigonometry identities, we know that:

sin²X + cos²X = 1

sin²X = 1 – cos²X = 1-(a/b)²

sin X = √(b²-a²)/b

sin A + sin²A = 1

sin A = 1 – sin²A

sin A = cos²A {since sin²θ + cos²θ = 1}

Squaring on both sides,

sin²A = (cos²A)²

1 – cos²A = cos⁴A

⇒ cos²A + cos⁴A = 1

We have,  
a cos θ – b sin θ = c  

Squaring both sides  
⇒ a²cos²θ + b²sin²θ – 2ab sin θ cos θ = c²  
⇒ a² (1 – sin²θ) + b² (1 – cos²θ) – 2ab sin θ cos θ = c²  
⇒ a² – a²sin²θ + b² – b²cos²θ – 2ab sin θ cos θ = c²  
⇒ a² + b² – c² = a²sin²θ + b²cos²θ + 2ab cos θ sin θ  
⇒ a² + b² – c² = (a sin θ + b cos θ)²  
⇒ (a sin θ + b cos θ) = ±√(a² + b² – c²)   → (1)  

So  
(a sin θ + b cos θ) – √(a² + b² – c²) = 0

⇒ acosθ + bsinθ = 4 --- (1)  
⇒ asinθ - bcosθ = 3 --- (2)  
→ Now, squaring and adding (1) and (2)  

∴ (acosθ + bsinθ)² + (asinθ − bcosθ)² = 4² + 3²  

⇒  
a²cos²θ + 2ab·sinθcosθ + b²sin²θ + a²sin²θ − 2ab·sinθcosθ + b²cos²θ = 16 + 9  

⇒ a²(sin²θ + cos²θ) + b²(sin²θ + cos²θ) = 25  

∴ a² + b² = 25 [∵ sin²x + cos²x = 1]