1(tan3A-tanA)-1(cot3A-cotA)=
tanA
tan2A
cotA
cot2A
Explanation for the correct answer:
Step 1. Simplify the given expression using trigonometric identities
1(tan3A-tanA)-1(cot3A-cotA)=1(tan3A-tanA)-11tan3A-1tanA [∵tanθ=1cotθ]
=1(tan3A-tanA)-tan3A.tanA(tanA-tan3A)
=1(tan3A-tanA)+tan3A.tanA(tan3A-tanA)
=1+tan3A.tanA(tan3A-tanA)
Step 2. Divide numerator and denominator by 1+tan3A.tanA
1(tan3A-tanA)-1(cot3A-cotA)=1tan3A-tanA1+tan3A.tanA
=1tan(3A-A) ∵tanA-tanB1+tanAtanB=tan(A-B)
=1tan2A=cot2A
Hence, Option ‘D’ is Correct.
loge(n+1)−loge(n−1)=4a[(1n)+(13n3)+(15n5)+...∞] Find 8a.