Question

Prove tanA+tan(90-A)=secA*sec(90-A)

Answer 1

tanA+tan(90°-A)
=tanA+cotA [∵, tan(90°-A)=cotA]
=tanA+1/tanA
=(tan²A+1)/tanA
=sec²A/tanA [∵, sec²A-tan²A=1]
=(1/cos²A)/(sinA/cosA)
=1/cos²A×cosA/sinA
=1/(cosAsinA)
=(1/cosA)(1/sinA)
=secAcosecA
=secAsec(90°-A) [∵, sec(90°-A)=cosecA] (Proved)

Answer 2

Answer:

To Prove : $tanA+tan(90-A)=secA \times sec(90-A)$

Solution :

$tanA+tan(90-A)$

Identity : $Tan(90-\theta )= Cot \theta$

$tanA+cot A$

Identity : $Cot \theta = \frac{1}{Tan \theta}$

$tanA+\frac{1}{tanA}$

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

Identity : $Sec^2 \theta - Tan^2\theta = 1$

$\frac{sec^2A}{tanA }$

Identity : $sec\theta = \frac{1}{cos \theta}, \frac{sin \theta }{cos \theta}=tan \theta$

$\frac{\frac{1}{cos^2A}}{\frac{Sin A}{Cos A}}$

$\frac{1}{(cosAsinA)}$

$\frac{1}{cos A}=sec A , \frac{1}{sinA}=cosec A$

$secAcosecA$

Using identity : $sec(90^{\circ}-A)=cosecA$

$secA \times sec(90-A)$

So, $tanA+tan(90-A)=secA \times sec(90-A)$

Hence proved