Question

if sina=ksin(a+b) show that tan (a+b) = sinb/cosb-k

Answer 1

Answer:

You are an ace so i think you know all the formula of compound angle so here is your answer

Step-by-step explanation:

Hope it helps you.

Answer 2

Answer:

$tan (a+b) = \frac{sinb}{cosb-k}$  is proved.

Step-by-step explanation:

Given:  $sina=ksin(a+b)$    $\implies$  $k=\frac{sina}{sin(a+b)}$    ......(i)

           $tan (a+b) = \frac{sinb}{cosb-k}$    .......(ii)

Taking RHS of eq (ii),

$\frac{sinb}{cosb-k}$

=  $\frac{sinb}{cosb-\frac{sina}{sin(a+b)} }$                  ....from (i)

= $\frac{sinb \hspace{.1cm} sin(a+b)}{cosb \hspace{.1cm} sin(a+b)-sina}$

= $\frac{sinb \hspace{.1cm} sin(a+b)}{cosb \hspace{.1cm}(sinacosb + cosasinb)-sina}$

= $\frac{sinb \hspace{.1cm} sin(a+b)}{sina \hspace{.1cm} cos^{2}b + cosa\hspace{.1cm} cosb \hspace{.1cm} sinb-sina}$

= $\frac{sinb \hspace{.1cm} sin(a+b)}{-sina \hspace{.1cm} (1- cos^{2}b) + cosa \hspace{.1cm} sinb\hspace{.1cm}cosb }$

= $\frac{sinb \hspace{.1cm} sin(a+b)}{-sina \hspace{.1cm} sin^{2}b+ cosa \hspace{.1cm} sinb\hspace{.1cm}cosb }$

= $\frac{sinb \hspace{.1cm} sin(a+b)}{sinb \hspace{.1cm} (cosa \hspace{.1cm} cosb - sina \hspace{.1cm} sinb)}$

= $\frac{ sin(a+b)}{cos(a+b) }$

= $tan(a+b)$

Hence proved.

#SPJ3