Question

show that secant theta minus tan theta whole square = 1_sin theta /1+ sin theta​

Answer 1

Answer:-

To Prove :-

(sec θ - tan θ)² = 1 - sin θ / 1 + sin θ

LHS:-

using -

• (a - b)² = a² + b² - 2ab

we get,

⟶ sec² θ + tan² θ - 2 sec θ tan θ

using sec θ = 1/cos θ and tan θ = sin θ / cos θ we get,

$\longrightarrow \sf \: \frac{1}{ \cos ^{2} \theta } + \frac{ { \sin }^{2} \theta}{ { \cos}^{2} \theta} - 2 \times \frac{1}{ \cos \theta} \times \frac{ \sin \theta }{ \cos \theta } \\ \\ \longrightarrow \sf \: \frac{1 + { \sin }^{2} \theta - 2 \sin \theta }{ { \cos}^{2} \theta}$

[ since , a² + b² - 2ab = (a - b)² where a = 1 , b = sec θ ]

we know that,

sin² θ + cos² θ = 1

⟶ cos² θ = 1 - sin² θ

So,

$\: \longrightarrow \sf \: \frac{(1 - \sin \theta) ^{2} }{1 - { \sin}^{2} \theta }$

• (a - b)² = (a - b)(a - b)

• a² - b² = (a + b)(a - b)

$\: \longrightarrow \sf \: \frac{(1 - \sin \theta) \cancel{(1 - \sin \theta)} }{(1 + \sin \theta) \cancel{(1 - \sin \theta)}} \\ \\ \longrightarrow \large\sf \frac{1 - \sin \theta }{1 + \sin \theta } = RHS$

Hence, Proved.

Answer 2

Given :

• secant theta minus tan theta whole square = 1_sin theta /1+ sin theta

$\text{To Prove : } \\ \\ \\ \sf \: \sec\theta-\tan\theta=\frac{1-\sin\theta}{1+\sin\theta} \: </p><p>$

Solution :

Taking R.H.S first :

$\sf : \implies\frac{1-\sin\theta}{1+\sin\theta}\\\\\text{Rationalizing the above fraction}\\\\ \sf : \implies \frac{1-\sin\theta}{1+\sin\theta}\times \frac{1-\sin\theta}{1-\sin\theta}\\\\ \sf : \implies \frac{(1-\sin\theta)^2}{1-\sin^2\theta}\\\\ \sf : \implies \frac{(1-\sin\theta)^2}{\cos^2\theta}\\\\ \sf : \implies(\frac{1-\sin\theta}{\cos\theta})^2\\\\ \sf : \implies (\frac{1}{\cos\theta}-\frac{\sin\theta}{\cos\theta})^2\\\\ \sf : \implies( \sf \sec\theta-\tan\theta)^2$

= R.H.S.

Hence Proved !!