Question

Find the value of $\sec AsinA+{tan}^{2}A-\csc A$ if $\tan A=2$ for $0\le A\le \frac{\pi }{2}$

Answer 1

Given : $\tan A=2$

$\therefore cotA=\frac{1}{tanA}=\frac{1}{2}$

$\because 1+co{t}^{2}A=cose{c}^{2}A$

$\Rightarrow cose{c}^{2}A=1+{\left(\frac{1}{2}\right)}^2=1+\frac{1}{4}=\frac{5}{4}$

$\Rightarrow cosecA=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}$

Now $secAsinA+ta{n}^{2}A-cosecA$

$\Rightarrow \frac{1}{cosA}sinA+ta{n}^{2}A-cosecA\because sec\theta =\frac{1}{cos\theta }$

$\Rightarrow tanA+ta{n}^{2}A-cosecA$ $\because tan\theta =\frac{sin\theta }{cos\theta }$

$\Rightarrow 2+2^2-\frac{\sqrt{5}}{2}$

$\Rightarrow 2+4-\frac{\sqrt{5}}{2}$

$\Rightarrow 6-\frac{\sqrt{5}}{2}=\frac{12-\sqrt{5}}{2}$

$\therefore$ The value of $secAsinA+ta{n}^{2}A-cosecA=\frac{12-\sqrt{5}}{2}$