tan ( 2 sin−1 ( 2 / √13 ) − 2 cos−1 ( 3 / √10 ) )
is equal to :
Correct Answer: 33/56
Let
α = sin−1( 2 / √13 )
Since sin−1x lies in the interval [−π/2, π/2], α lies in the first quadrant.
So we take a right-angled triangle with:
sin α = 2 / √13
Hence,
cos α = √(1 − sin²α) = √(1 − 4/13) = 3 / √13
Now,
tan α = (2 / √13) / (3 / √13) = 2/3
Using the identity:
tan 2α = 2 tan α / (1 − tan² α)
we get:
tan 2α = 2(2/3) / (1 − 4/9) = (4/3) / (5/9) = 12/5
Now let
β = cos−1( 3 / √10 )
Since cos−1x lies in [0, π], β lies in the first quadrant.
So,
sin β = √(1 − cos²β) = √(1 − 9/10) = 1 / √10
Hence,
tan β = (1 / √10) / (3 / √10) = 1/3
Using the identity:
tan 2β = 2 tan β / (1 − tan² β)
we get:
tan 2β = 2(1/3) / (1 − 1/9) = (2/3) / (8/9) = 3/4
Now the required expression is:
tan(2α − 2β)
Using the identity:
tan(A − B) = (tan A − tan B) / (1 + tan A tan B)
Substitute tan 2α = 12/5 and tan 2β = 3/4:
tan(2α − 2β) = (12/5 − 3/4) / (1 + (12/5)(3/4))
= (48 − 15)/20 / (1 + 36/20)
= (33/20) / (56/20) = 33/56
Hence, the value of the given expression is
33/56
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.