Question

is there any other method to solve this sum

Answer 1

Yes! There is a method dear, I'm gonna solve this question for you!

Answer:

Numerator = tan theta + sec theta -1

= sin theta/cos theta + 1/cos theta - 1

= (sin theta + 1)/cos theta -1

denominator = (sin theta -1)/cos theta +1

multiply both by cos theta

numerator = sin theta +1 - cos theta

= 2 sin theta/2 cos theta/2 + 2 sin^ theta/2

denominator = 2 sin theta/2 cos theta/2 - 2 sin ^2 theta/2

divide

number = (cos theta/2 + sin theta/2)/(cos theta/2 - sin theta/2)

= (cos theta/2 + sin theta/2)^2/(cos^2 theta/2- sin ^2 theta/2)

= (cos^2 theta/2+ sin^2 theta/2 + 2 cos theta/2 sin theta/2)/(cos theta)

= (1+ sin theta)/ cos theta

= sec theta+ tan theta

Hence Proved

I Hope It Will Help!

^_^

Answer 2

Step-by-step explanation:

Note: It can be solved in many methods. For the better understanding I am replacing θ with A.

Important Formulas:

(i) sec²θ - tan²θ = 1

(ii) sec²θ = 1 + tan²θ

(iii) tan²θ = sec²θ - 1

Now,

$Given:\frac{tanA+secA-1}{tanA-secA+1}$

$=\frac{tanA+(secA-1)}{tanA-(secA-1)}*\frac{tanA+(secA-1)}{tanA+(secA-1)}$

$=\frac{[tanA+(secA-1)]^2}{(tanA)^2-(secA-1)^2}$

$=\frac{tan^2A + sec^2A + 1 - 2secA + 2tanAsecA-2tanA}{tan^2A-sec^2A-1+2secA}$

$=\frac{2sec^2A-2secA+2tanAsecA-2tanA}{tan^2A-1-tan^2A-1+2secA}$

$=\frac{2sec^2A-2secA+2tanAsecA-2tanA}{2secA-2}$

$=\frac{sec^2A-secA+tanAsecA-tanA}{secA-1}$

$=\frac{secA(secA - 1) + tanA(secA - 1)}{(secA - 1)}$

$=\frac{(secA+tanA)(secA - 1)}{secA-1}$

$=secA + tanA$

$=\frac{1}{cosA} + \frac{sinA}{cosA}$

$=\boxed{\frac{1+sinA}{cosA}}$

Hope it helps!