If 2tanA=3tanB, then sin2B5-cos2B is equal to
tanA–tanB
tan(A–B)
tan(A+B)
tan(A+2B)
Explanation for the correct option:
Step 1. Find the value of sin2B5-cos2B:
Given, 2tanA=3tanB ….(1)
sin2B5-cos2B=2tanB1+tan2B5–1–tan2B1+tan2B ∵sin2θ=2tan(θ2)1+tan2(θ2),cos2θ=1-tan2(θ2)1+tan2(θ2)
=2tanB5(1+tan2B)–1+tan2B=2tanB5+5tan2B–1+tan2B=2tanB4+6tan2B=2tanB22+3tan2B=tanB2+3tanBtanB
Step 2. Divide the numerator and denominator by 2:
⇒12×tanB1+32×tanB×tanB
⇒32×tanB–tanB1+tanAtanB [from equation(1)]
⇒tanA–tanB1+tanAtanB
⇒tan(A–B)
Hence, Option ‘B’ is Correct.
If 5a+2b5a−2b=53, then a : b =