Question

साइन थीटा माइनस कोस थीटा प्लस वन अपॉन साइन थीटासाइन थीटा माइनस कोस थीटा प्लस वन अपॉन साइन थीटा प्लस कोस थीटा माइनस वन इज इक्वल टू वन अपॉन सेक्स ​

Answer 1

Answer:

let theta = a

(sin a - cos a +1) / (sin a + cos a -1) = 1/ sec a - tan a

LET LHS

(sin a - cos a +1) / (sin a + cos a -1)

NUMERATOR AND DENOMINATOR DIVIDED BY cos a

then

(tan a -1 + sec a) / (tan a + 1 - sec a)

(tan a + sec a -1) / (tan a + 1 - sec a)

[tan a + sec a - ( sec^ 2a - tan^2a ) ] /(tan a +1 - sec a)

(tan a+ sec a) { 1 - sec a + tan a)} / ( tan a +1 - sec a)

tana + sec a

now multiply and divided by )(sec a - tan a)

then

(sec a + tan a) ( sec a- tan a) /

( sec a- tan a)

(sec^ 2 a - tan ^2 a) / (sec a- tan a)

1 / sec a - tan a RHS

PLEASE MARK AS A BRAINILIST

Answer 2

Given: $\[\frac{{\left( {\sin \theta - \cos \theta + 1} \right)}}{{\sin \theta - \cos \theta - 1}} = \frac{1}{{\sec \theta - \tan \theta }}\]$.

To show: Whether LHS is equal to RHS.

Solution:

Take LHS and Solve it.

$\[LHS = \frac{{\left( {\sin \theta - \cos \theta + 1} \right)}}{{\sin \theta - \cos \theta - 1}}\]$

$\[ = \left( {\frac{{\left( {\sin \theta - \cos \theta + 1} \right)}}{{\sin \theta - \cos \theta - 1}}} \right)\left( {\frac{{\left( {\sin \theta + \cos \theta + 1} \right)}}{{\sin \theta - \cos \theta - 1}}} \right)\]$

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

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

$\[ = \frac{{1 + \sin \theta }}{{\cos \theta }}\]$

$\[ = \left( {\frac{{1 + \sin \theta }}{{\cos \theta }}} \right)\left( {\frac{{1 - \sin \theta }}{{1 - \sin \theta }}} \right)\]$

$\[ = \frac{{\cos \theta }}{{1 - \sin \theta }}\]$

$\[\begin{array}{l} = \frac{1}{{\sec \theta - \tan \theta }}\\ = RHS\end{array}\]$

Hence proved.