JEE Advanced Level Test: Trigonometric Ratio- 1 - NDA MCQ

 tan1° tan2° .........tan89°
= (tan1° tan89°)(tan2° tan85°)........tan45°
= (tan1° cot1°)(tan2° cot°)......tan45°
= 1•1 = 1

{tan(x-π/2).cos(3π/2+x)-sin³(7π/2-x)}/{cos(x-π/2)tan(3π/2+x)}
= [tan{-(π/2-x)}.cos{(π/2×3)+x} - sin³{(π/2×7)-x}]/[cos{-(π/2-x)}.tan{(π/2×3)+x}
= [-tan(π/2-x).sinx - (-cosx)³]/[cos(π/2-x).(-cotx)]
= (-cotx.sinx+cos³x)/[sinx.(-cotx)]
= [-(cosx/sinx).sinx+cos³x]/[sinx.(-cosx/sinx)]
= {-cosx(1 - cos²x)}/(-cosx)
= 1 - cos²x
= sin²x

Correct Answer :- d

Explanation : sin(p + q) sin(p -q)cosec²q

= (sin²p –sin²q)1/sin²q

= sin²p/sin²q - 1

sin A sin B - cos A cos B + 1 = 0
=> cos A cos B -sin A sin B = 1
⇒ cos(A+B) = 1 = cos0
⇒ A = 2nΠ − B , where n∈Z
Now 1+cotAtanB
⇒ 1+cot(2nπ−B)tanB
⇒ 1 − cot(B)tanB
= 1−1   = 0

3A= A+ 2A
⇒ tan 3A = tan (A + 2A)
⇒ tan 3 A = tanA + tan2A/ 1 – tan A . tan 2A
⇒ tan A + tan 2A = tan 3A – tan 3A. tan 2A . tan A
⇒ tan 3 A – tan 2A – tan A = tan 3A . tan 2A . tan A

tan225=+1
tan(203+22)=+1
= tan203+tan22/1−tan203tan22 = +1
⇒ tan203+tan22=+1−tan203tan22
⇒ tan203+tan22+tan203tan22=1

A lies in third quadrant = 180degree < A ≤270degree
3tan A - 4 = 0
tan A = 4/3 
sinA = -4/5 ; cos A = -3/5 
A = 5sin(2A) + 3sinA + 4cosA
= 10(-4/5)(-3/5) + 3(4/5) + 4(-3/5)
 = 0

So, (cos20+8sin10sin50sin70)/sin²(80)
= cos20+8(1/4*sin³(10))/sin²(80)
= cos20+1/sin²(80)
= 2cos²(10)/sin²(80)
= 2sin²(80)/sin²(80)
= 2

sin 12°sin 48°sin 54°

α+β+γ=2π
⇒(α+β+γ)/2 =π
⇒ [tanα/2 + tanβ/2+  tanγ/2 − tanα/2 tanβ/2 tanγ/2]/[1 −tanα/2 tanβ/2 − tanβ/2 tanγ/2 - −tanα/2 tanγ/2] = 0
= tanα/2 tanβ/2 tanγ/2

AC = x/(sin∠ACD) = x/sinC
AB = x/sinB
x²(1/sin²C + 1/sin²B) = 8x²
sin²B + sin²C = 8sin²C sin²B

1 = 8sin²C cos²C
(sin²C) = 1/√2 ⇒ C = π/8 ⇒ B = 3π/8