Question

1-SinA/1+sinA=(seca-tana) ^2​

Answer 1

$Answer \: \\ \\ ( \sec(a) - \tan(a) ) {}^{2} = \frac{1 - \sin(a) }{1 + \sin(a) } \\ \\ lhs \\ \\ ( \frac{1}{ \cos(a) } - \frac{ \sin(a) }{ \cos(a) } ) {}^{2} \\ \\ \frac{(1 - \sin(a)) {}^{2} }{ \cos {}^{2} (a) } \\ \\ \frac{(1 - \sin(a) ) {}^{2} }{1 - \sin {}^{2} (a) } \\ \\ \frac{(1 - \sin(a) ) \times (1 - \sin(a) )}{(1 - \sin(a)) \times (1 + \sin(a) ) } \\ \\ \frac{1 - \sin(a) }{1 + \sin(a)} \: \: \: hence \: proved \: \\ \\ NOTE \\ \: \\ 1) \: \: \: \sec(x) = \frac{1}{ \cos(x) } \\ \\ 2) \: \: \: \tan(x) = \frac{ \sin(x) }{ \cos(x) } \\ \\ 3) \: \: \: sin {}^{2} (x) = 1 - \cos {}^{2} (x) \\ \\ 4) \: \: \: ( \alpha {}^{2} - \beta {}^{2} ) = ( \alpha - \beta ) \times ( \alpha + \beta )$

Answer 2

Step-by-step explanation:

$\begin{lgathered}Answer \: \\ \\ ( \sec(a) - \tan(a) ) {}^{2} = \frac{1 - \sin(a) }{1 + \sin(a) } \\ \\ lhs \\ \\ ( \frac{1}{ \cos(a) } - \frac{ \sin(a) }{ \cos(a) } ) {}^{2} \\ \\ \frac{(1 - \sin(a)) {}^{2} }{ \cos {}^{2} (a) } \\ \\ \frac{(1 - \sin(a) ) {}^{2} }{1 - \sin {}^{2} (a) } \\ \\ \frac{(1 - \sin(a) ) \times (1 - \sin(a) )}{(1 - \sin(a)) \times (1 + \sin(a) ) } \\ \\ \frac{1 - \sin(a) }{1 + \sin(a)} \: \: \: hence \: proved \: \\ \\ NOTE \\ \: \\ 1) \: \: \: \sec(x) = \frac{1}{ \cos(x) } \\ \\ 2) \: \: \: \tan(x) = \frac{ \sin(x) }{ \cos(x) } \\ \\ 3) \: \: \: sin {}^{2} (x) = 1 - \cos {}^{2} (x) \\ \\ 4) \: \: \: ( \alpha {}^{2} - \beta {}^{2} ) = ( \alpha - \beta ) \times ( \alpha + \beta )\end{lgathered}$

Answer

(sec(a)−tan(a))

2

=

1+sin(a)

1−sin(a)

lhs

(

cos(a)

1

−

cos(a)

sin(a)

)

2

cos

2

(a)

(1−sin(a))

2

1−sin

2

(a)

(1−sin(a))

2

(1−sin(a))×(1+sin(a))

(1−sin(a))×(1−sin(a))

1+sin(a)

1−sin(a)

henceproved

NOTE

1)sec(x)=

cos(x)

1

2)tan(x)=

cos(x)

sin(x)

3)sin

2

(x)=1−cos

2

(x)

4)(α

2

−β

2

)=(α−β)×(α+β)