Question

Prove tan theta + sec theta minus one upon tan theta minus sec theta + 1 is equals to 1 + sin theta upon cos theta

Answer 1

Solution:

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

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

Hence proved.

Answer 2

"$\frac { \tan { \left( \theta \quad +\quad \sec { \left( \theta -1\right)} \right)}}{ \tan { \left( \theta \quad -\quad \sec { \left( \theta +1\right)}\right)}} \quad =\quad \frac {1\quad +\quad \sin { \theta}}{ \cos {\theta}}$

LHS:

$=\quad \frac { \left( \tan { \theta} \quad +\quad \sec {\theta}\right)\quad -\quad \left( \sec ^{ 2 }{ \theta} \quad -\quad \tan ^{ 2 }{\theta}\right)}{ \tan {\theta} \quad -\quad \sec {\theta} \quad +\quad 1}$

$=\frac { \left( \tan {\theta} \quad +\quad \sec {\theta}\right) \quad -\quad \left( \sec { \theta} \quad +\quad \tan {\theta} \right) \left( \sec {\theta} \quad -\quad \tan { \theta}\right)}{ \tan { \theta} \quad -\quad \sec { \theta} \quad +\quad 1}$

$=\quad \frac { \left( \tan {\theta} \quad +\quad \sec {\theta} \right) \left[ 1\quad -\quad \sec {\theta} \quad +\quad \tan {\theta} \right]}{\tan {\theta} \quad -\quad \sec {\theta} \quad +\quad 1 }$

$=\quad \tan {\theta} \quad +\quad \sec {\theta}$

$=\quad \frac { \left( \tan { \theta} \quad +\quad \sec { \theta}\right) \left( \sec { \theta} \quad -\quad \tan { \theta}\right)  }{ \left( \sec { \theta} \quad -\quad \tan { \theta} \right)}$

$=\quad \frac { \sec ^{ 2 }{ \theta} \quad +\quad \tan ^{ 2 }{ \theta}}{ \sec { \theta} \quad -\quad \tan { \theta}} \\ =\quad \frac { 1 }{ \sec { \theta } \quad -\quad \tan { \theta}}$

$=\quad \frac { 1 }{ \frac { 1 }{ \cos { \theta}} \quad -\quad \frac { \sin { \theta}}{ \cos { \theta}}}$

$=\quad \frac { 1 }{ \frac { 1\quad -\quad \sin { \theta}}{\cos { \theta}}}$

$=\quad \frac { \cos { \theta}}{ 1\quad -\quad \sin { \theta}}$

$=\quad \frac { \cos { \theta}}{ 1\quad -\quad \sin {\theta}} \quad \times \quad \frac {1\quad +\quad \sin { \theta}}{ 1\quad +\quad \sin { \theta}}$

$=\quad \frac { \cos { \theta} \quad \left( 1\quad +\quad \sin { \theta} \right)}{ 1\quad -\quad \sin ^{ 2 }{ \theta}} \\ =\quad \frac { \cos { \theta} \quad \left( 1\quad +\quad \sin {\theta} \right)}{ \cos ^{ 2 }{ \theta} }$

$=\quad \frac { 1\quad +\quad \sin {\theta}}{ \cos { \theta}}$

$=\quad RHS$

Hence proved."