Lakshman Patel RJIT
asked
in Quantitative Aptitude
Apr 1, 2020
recategorized
Nov 8, 2020
by Krithiga2101

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Ans is option (C)

Substitute $cotA=\frac{1}{tanA}$ in the above question, we get

$\frac{tan^{2}A}{tanA-1}+\frac{1}{tanA(1-tanA)}$

Simplifying further we get, $\frac{1}{tanA-1}[tan^{2}A-\frac{1}{tanA}]$

$\Rightarrow$ $\frac{tan^{3}A-1}{(tanA-1)tanA}$

$\Rightarrow$ $\frac{(tanA-1)(tan^{2}A+tanA+1)}{(tanA-1)tanA}$ (using $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$)

$\Rightarrow$ $tanA+cotA+1$

Substitute $cotA=\frac{1}{tanA}$ in the above question, we get

$\frac{tan^{2}A}{tanA-1}+\frac{1}{tanA(1-tanA)}$

Simplifying further we get, $\frac{1}{tanA-1}[tan^{2}A-\frac{1}{tanA}]$

$\Rightarrow$ $\frac{tan^{3}A-1}{(tanA-1)tanA}$

$\Rightarrow$ $\frac{(tanA-1)(tan^{2}A+tanA+1)}{(tanA-1)tanA}$ (using $a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})$)

$\Rightarrow$ $tanA+cotA+1$