Question

If alpha + beta =90 degrees, then proof that (sec alpha +sin beta) /sin alpha =tan alpha +2 tan beta. Plz give me the correct answer. It is urgent

Answer 1

$\alpha + \beta = 90 \\ \frac{ \sec( \alpha ) + \sin( \beta ) }{ \sin( \alpha ) } = \tan( \alpha ) + 2 \tan( \beta ) \\ rhs \\ = \tan( \alpha ) + 2 \tan( \beta ) \\ = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } + 2\frac{ \sin( \beta ) }{ \cos( \beta ) } \\ = \frac{ \sin( \alpha ) \cos( \beta ) + 2 \sin( \beta ) \cos( \alpha ) }{ \cos( \alpha ) \cos( \beta ) } \\ = \frac{ \sin( \alpha ) \cos(90 - \alpha ) + 2 \sin( \beta ) \cos( \alpha )}{ \cos( \alpha ) \cos(90 - \alpha ) } \\ = \frac{ \sin( \alpha ) \sin( \alpha ) + 2 \sin( \beta ) \ \cos( \alpha ) }{ \cos( \alpha ) \sin( \alpha ) } \\ = \frac{1 - { \cos( \alpha ) }^{2} + 2 \sin( \beta ) \cos( \alpha ) }{ \cos( \alpha ) \sin( \alpha ) } \\ = \frac{1}{ \cos( \alpha ) \sin( \alpha ) } + \frac{ - \cos( \alpha ) + 2 \sin( \beta ) }{ \sin( \alpha ) } \\ = \frac{1}{ \cos( \alpha ) \sin( \alpha ) } + \frac{ - \cos(90 - \beta ) + 2 \sin( \beta ) }{ \sin( \alpha ) } \\ = \frac{1}{ \cos( \alpha ) \sin( \alpha ) } + \frac{ \sin( \beta ) }{ \sin( \alpha ) } \\ = \frac{ \sec( \alpha ) }{ \sin( \alpha ) } + \frac{ \sin( \beta ) }{ \sin( \alpha ) } \\ = \frac{ \sec( \alpha ) + \sin( \beta ) }{ \sin( \alpha ) } \\ = lhs \\ proved$

Answer 2

it is true when α + β = 90° then (sec α +sin β) /sin α = tan α +2 tan β  

Given:

α + β = 90°

(sec α +sin β) /sin α = tan α +2 tan β

To find:

Prove that (sec α +sin β) /sin α = tan α +2 tan β

Solution:

Given α + β = 90°

We need to prove (sec α +sin β) /sin α = tan α +2 tan β  

Take  RHS =  tan α +2 tan β  

= $\frac{sin \alpha }{cos\alpha } + \frac{2 sin\beta }{cos \beta }$

= $\frac{sin\alpha cos\beta +2 sin\beta cos\alpha }{ cos\alpha cos \beta }$

= $\frac{sin\alpha cos(90 - \alpha ) +2 sin\beta cos\alpha }{ cos\alpha cos (90-\alpha ) }$   [ ∵ β = 90 - α ]

= $\frac{sin\alpha sin \alpha +2 sin\beta cos\alpha }{ cos\alpha sin \alpha }$

= $\frac{sin^{2} \alpha +2 sin\beta cos\alpha }{ cos\alpha sin \alpha }$

= $\frac{1 - cos^{2} \alpha +2 sin\beta cos\alpha }{ cos\alpha sin\alpha }$           [ ∵ sin²α = 1 - cos²α]

= $\frac{1}{cos\alpha sin\alpha} + \frac{- cos^{2} \alpha +2 sin\beta cos\alpha }{ cos\alpha sin\alpha }$  

=  $\frac{1}{cos\alpha sin\alpha} + \frac{cos\alpha (-cos\alpha +2 sin\beta ) }{ cos\alpha sin\alpha }$

= $\frac{1}{cos\alpha sin\alpha} + \frac{ (-cos\alpha +2 sin\beta ) }{ sin\alpha }$

= $\frac{1}{cos\alpha sin\alpha} + \frac{ (- sib\beta +2 sin\beta ) }{ sin\alpha }$

= $\frac{1}{cos\alpha sin\alpha} + \frac{ ( sin\beta ) }{ sin\alpha }$

=  $\frac{sec\alpha }{sin\alpha} + \frac{ sin\beta }{ sin\alpha }$  

=  $\frac{ sec\alpha+sin\beta }{ sin\alpha }$ = LHS

Therefore, it is proven that (sec α + sin β) /sin α = tan α +2 tan β  

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