Question

(ii) If tan theta
12/5
then find the value of
1 + sin theta / 1 - sin thrta


​

Answer 1

Solution:

$\tt{tan \theta = \dfrac{12}{5}}$

$\leadsto \tt{\dfrac{Opposite \: side}{Adjacent \: side} = \dfrac{12}{5}}$

So,

• Opposite side = 12 cm
• Adjacent side = 5 cm

By Pythagoras Theorem,

(AC)² = AB² + BC²

⇒ AC² = 12² + 5²

⇒ AC² = 144 + 25

⇒ AC²= 169

⇒ AC = √169

⇒ AC = 13 cm

$\sf{sin \theta = \dfrac{Opposite \:side}{Hypotenuse} = \dfrac{AB}{AC} = \dfrac{12}{13}}$

$\sf{\dfrac{1 + sin \theta}{1 - sin\theta}= \dfrac{1 + \dfrac{12}{13}}{1 - \dfrac{12}{13}}= \dfrac{\dfrac{25}{13}}{\dfrac{1}{13}}= \dfrac{25}{13} \times \dfrac{13}{1} = 25}$

∴ 1 + sinθ / 1 - sinθ = 25

Answer 2

Given :

$\\ \qquad \bull \: \: \underline{ \boxed {\tt tan \: \theta \: \implies \: \dfrac{12}{5} }} \: \: \large\star$

To Find :

$\\ \\ \qquad \bull \: \: \underline{ \boxed{ \tt \: \frac{1 + sin \: \theta}{1 - sin \: \theta} }} \: \: \large \star$

Solution :

From the above figure,

$➠ \: \: \tt \: AC \: = \: \sqrt{(AB) {}^{2} + (BC) {}^{2} }$

$➠ \: \tt\sqrt{12 {}^{2} \: + \: 5 {}^{2} }$

$➠\: \tt\sqrt{144+5^{2}}$

$➠ \: \tt\sqrt{144+25}$

$➠ \: \tt\sqrt{169}$

$\underline{ \boxed { \color{purple}{ \frak{➯ \:13 }}}} \: \: \bigstar$

Then,

$: \implies \tt sin \: \theta \: = \: \dfrac{12}{13}$

Therefore,

$: \implies \tt \: \dfrac{1 + sin \: \theta}{1 - sin \: \theta} \: ➠ \: \dfrac{1 + \frac{12}{13} }{1 - \frac{12}{13}}$

$➯\:\tt\dfrac{\frac{13}{13}+\frac{12}{13}}{1-\frac{12}{13}}$

$➯\:\tt\dfrac{\frac{13+12}{13}}{1-\frac{12}{13}}$

$➯\:\tt\dfrac{\frac{13+12}{13}}{1-\frac{12}{13}}$

$➯\:\tt\dfrac{\frac{25}{13}}{\frac{13}{13}-\frac{12}{13}}$

$➯\:\tt\dfrac{\frac{25}{13}}{\frac{13-12}{13}}$

$➯\:\tt\dfrac{\frac{25}{13}}{\frac{1}{13}}$

$➯\:\tt\dfrac{25}{ \cancel{13}}\times \cancel{13 }$

$\underline{ \boxed { \color{purple}{ \frak{➠ \:25 }}}} \: \: \bigstar$

.°. Hence, the value of the given expression is 25.

✰ Hope it helps ✰