Question

1-sin theta / cos theta + cos theta / 1- sin theta = 2 sec theta​

Answer 1

Sᴏʟᴜᴛɪᴏɴ :-

Taking LHS ,

→ (1 - sinA)/cosA + cosA/(1 - sinA)

Taking LCM ,

→ {(1 - sinA)² + cos²A}/cosA.(1 - sinA)

using (a - b)² = (a² + b² - 2ab) we get,

→ {1+sin²A - 2sinA + cos²A}/cosA.(1 - sinA)

using sin²A + cos²A = 1 in numerator now,

→ (1 + 1 - 2sinA)/cosA(1 - sinA)

→ (2 - 2sinA)/cosA(1 - sinA)

Taking 2 common from numerator now,

→ 2(1 - sinA)/cosA(1 - sinA)

→ 2/cosA

Putting (1/cosA) = secA in Last ,

→ 2secA = RHS (Proved).

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\dagger \: {\sf{ \frac{1 - sin \theta}{cos \theta} + \frac{cos \theta}{1 - sin \theta} = 2sec \theta \: }}$

${\bf{\blue{\underline{Now:}}}}$

Take L.H.S,

${\sf{ L.H.S = \frac{1 - sin \theta}{ \cos \theta} + \frac{cos \theta}{1 - sin \theta} }}$

${\sf{ = \frac{(1 - sin \theta)(1 - sin \theta) + (cos \theta)(cos \theta }{(cos \theta )\:(1 - sin \theta) } }}$

${\sf{ = \frac{( {1 - sin \theta) ^{2} + {cos}^{2} \theta } }{cos \theta \: (1 - sin \theta)} }}$

$\dagger \boxed{\sf{ \purple{(x - y) ^{2} = {x}^{2} + {y}^{2} - 2xy} }}$

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

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

$\dagger \boxed{\sf{ \purple{cos ^{2} \theta + {sin}^{2} \theta = 1} }}$

${\sf{ = \frac{ {1} + 1 - 2sin \theta }{cos \theta(1 - sin \theta)} }}$

${\sf{ = \frac{ 2- 2sin \theta }{cos \theta(1 - sin \theta)} }}$

${\sf{ = \frac{ 2(1- sin \theta) }{cos \theta(1 - sin \theta)} }}$

${\sf{ = \frac{ 2 \cancel{(1- sin \theta)} }{cos \theta \cancel{(1 - sin \theta)}} }}$

${\sf{ = \frac{ 2 }{cos \theta } }}$

${\sf{ = 2sec\theta}}$

Hence L.H.S = R.H.S