Question

Prove the following identities tan theta + sec theta minus one upon tan theta minus sec theta + 1 is equal to 1 + sin theta upon cos theta​

Answer 1

Solution:

Used the trigonometric identity: sec²A- tan²A=1, to solve the problem.

\begin{lgathered}=\frac{tan A + sec A -1}{tan A - sec A +1}\\\\ =\frac{(tan A + sec A) -(Sec^2A-tan^2A)}{tan A - sec A +1}\\\\ =\frac{(tan A + sec A)[1-(sec A-tan A)]}{tan A - sec A +1}\\\\=\frac{(tan A + sec A)[1-sec A+tan A)]}{tan A - sec A +1}\\\\= tan A + sec A\\\\= \frac{sin A}{cosA}+\frac{1}{cosA}\\\\ =\frac{1 +sinA}{cosA}\end{lgathered}

=

tanA−secA+1

tanA+secA−1

=

tanA−secA+1

(tanA+secA)−(Sec

2

A−tan

2

A)

=

tanA−secA+1

(tanA+secA)[1−(secA−tanA)]

=

tanA−secA+1

(tanA+secA)[1−secA+tanA)]

=tanA+secA

=

cosA

sinA

+

cosA

1

=

cosA

1+sinA

Hence proved.