JEE Advanced Level Test: Fundamentals of Trigonometric - NDA MCQ

(cot α₁) (cot α₂)........... cot α_n =1
(cos α₁) (cos α₂) .......... cos α_n = (sin α₁) (sin α₂) .......... sin α_n
Let y = (cos α₁) (cos α₂) ................ (cos α_n) (to be maximum)
Squaring y² = (cos² α₁) (cos² α₂) ................ (cos² α_n)
= (cos α₁ sin α₁) (cos α₂ sin α₂) ................ (cos α_n sin α_n)    using equation 1

[sin 2α₁. sin2α₂ .........sin 2α_n]
0 < sin2α₁. sin2α₂ .........sin 2α_n < 1

We know, < a sinx + bcos x 
∴ 7cosx+5sinx
 
Since k is integer,   -9 < 2k + 1 < 9 ⇒ -10 < 2k < 8 ⇒ -5 < k < 4
⇒ Number of possible integer values of k = 8.

sin³10° + sin³50° - sin³70° 
= 1/4 [(3 sin 10° - sin 30°) + (3 sin 50° - sin 150°) - (3 sin 70° - sin 210°)]

put x = tan A, y = tan B, z = tan C
tan A tan B + tan B tan C + tan C tan A = 1
tan C [tan A + tan B] = 1– tan A tan B

since n is appositive integer

sinθ+sin3θ+sin2θ=sinα
⇒ 2 sin 2θ cosθ + sin 2θ = sin α

⇒ sin 2θ(2 cos θ +1) = sin α........(1)
Now cos θ + cos 3θ + cos 2θ = cos α cos 2θ (2 cos θ + 1) = cos α. ....(2)
From (1) and (2), tan 2θ = tan α ⇒ 2θ = α ⇒θ = α/2 

So, sin (β - α) = sin β.cosα - cos β.sin a

3[cos⁴ α + sin⁴ α] - 2[cos⁶ α + sin⁶ α]
= 3[(cos² α + sin² α)² - 2sin²α.cos² α]- 2
[(sin² α + cos² α)³ - 3 sin² α.cos² α.(sin² α + cos² a)]
= 3[1 - 2sin² α.cos² α] - 2[1 - 3 sin² α.cos² α] = 3 - 2 = 1

The given equation can be written as 

sin A = sin B ⇒ sin A - sim B = 0
...........(1)
and cos A = cos B ⇒ cos B - cos A = 0
    ..........(2)

Equations (1) and (2) are simultaneously true if sin (1/2) (A – (B) = 0,
while the other factors sin (1/2) (A+ (B) and cos (1/2) (A + (B) cannot both be zero simultaneously. 

In a ΔABC, we know that
tanA + tanB + tanC = tanA tanB tanC
∴ tanB + tanC = tanA(tanBtanC – 1) 


⇒ tan² B - (K - 1) tan B + K = 0
For real values of tan B, Disc.
(K – 1)² – 4K > 0 
K² – 6K + 1 > 0 

sin x + sin y = sin (x + y)




The possibilities are x = 0 or, y = 0 or, x + y = 0
When x = 0 , y = ±1
When y = 0 , x = ±1

Clearly, there are 6 pairs.

Put θ = A, 2θ = B, 4θ = c
∴ A + B + C = 7θ = 2π

But cos (A + B + C)
= cos A cos B cos C - ∑ sin A sin B cosC
∴ sin A sin Bsin C = cos A cos Bcos C - cos 2π

sin² (θ - α) cos α = cos² (θ - α) sin α = m sin α cos α

cos10⁰.cos 20⁰.cos 40⁰

cos 25⁰ + sin 25⁰ = K
Squaring,
1 + sin 50⁰ = K²
sin 50⁰ = K-1

sin x + cosec x > 2 (using A.M > G.M)
Also tan y + cot > 2
So, sin x + cosec x + tan y + cot y = 4 is possible when 




So, tan y/2 is a root of the equation α2 + 2α - 1 = 0

⇒ cot B + tan A = 2 tan A – 2 tan B
⇒ 2 tan B + cot B = tan A