Question

ques no. 26 prove that

Answer 1

Step-by-step explanation:

LHS

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

We know that sec²Ф - tan²Ф = 1, putting it in Numerator in place of 1

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

Using a² - b² = ( a + b ) (a - b)

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

Taking (sec A + tan A) common from numerator

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

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

$=secA+tanA$

∵ secФ = 1/cosФ and tanФ = sinФ/cosФ, therefore

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

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

=RHS  Hence Proved