Math Olympiad Test: Trigonometry- 3 - Free MCQ with solutions Class 10

We have, 

(sin α + cosec α)² + (cos α + sec α)²  = k + tan²α + cot² α
⇒ sin²α + cosec²α + 2 sin α . cosec α + cos²α + sec²α + 2 cos α.sec α = k + tan²α + cot²α
⇒ sin²α + cos²α + 2 + cosec²α + sec²α + 2 = k + tan²α + cot²α
⇒ 5 + cosec²α + sec²α = k + tan²α + cot²α
⇒ 5 + 1 + cot²α + 1 + tan²α = k + tan²α + cot²α
⇒ 7 + cot²α + tan²α = k + tan²α + cot²α
⇒ k = 7

We have, A + B = 90° ⇒ A = 90° – B ...(i)
Now, 
 [using (i)]

We have, sinθ + cosθ = a and secθ + cosecθ = b
Now, b(a² – 1) =   (sin²θ + cos²θ + 2sinθcosθ – 1)

acosθ + bsinθ = m
Squaring both sides, we get
a²cos²θ + b²sin²θ + 2ab cosθsinθ = m² ... (i)
a sinθ – b cosθ = n
Squaring both sides, we get
a²sin²θ + b²cos²θ – 2ab cosθsinθ = n² ... (ii)
Adding (i) and (ii), we get
a² + b² = m² + n²

sinθ = cosθ ⇒ sinθ/cosθ = 1 ⇒ tanθ = 1 and tanθ = tan 45° ⇒ θ = 45°
∴ 2 tan²θ + sin²θ – 1 = 2 tan² 45° + sin² 45° – 1

We have, sin(A + B + C) = 1
⇒ sin(A + B + C) = sin 90°
⇒ A + B + C = 90° ... (i)
Also, tan(A – B) = 1/√3 = tan 30°
⇒ A – B = 30° ... (ii)
and sec (A + C) = 2 = sec 60°
⇒ A + C = 60° ... (iii)
From (ii) and (iii), we get
B + C = 30° ... (iv)
From (i) and (iv), we get, A = 60°
∴ B = 30° [Using A = 60° in (ii)]
and C = 0° [Using A = 60° in (iii)]

We have, (1 + tan² A) +
= sec² A + (1 + cot² A) [∵ 1 + tan² A = sec²
A]
= sec² A + cosec² A [∵ 1 + cot² A = cosec²
A]

x = r sinθcosφ ... (i)
y = r sinθsinφ ... (ii)
z = r cosθ ... (iii)
Squaring and adding (i) and (ii), we get
x² + y² = r²sin²q ... (iv)
Squaring (iii) and adding it with (iv), we get
x² + z² + y² = r²