If the numerical value of tan(cos−1(4/5)+tan−1(2/3)) is a/b then
A
a+b=23
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B
a−b=11
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C
3b=a+1
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D
2a=3b
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Solution
The correct options are Aa+b=23 Ba−b=11 C3b=a+1 Let cos−1(4/5)=α, that is, cosα=4/5, so that tanα=√(54)2−1=34(∵0α<πandcosα>0) and tan(cos−145+tan−123)=tanα+231−tanα23 =34+231−23⋅34=176=ab (Given) So a=17,b=6,a+b=23,a−b=11 and 3b=a+1