Question

pls answer me friends​

Answer 1

Answer:

(B) nambar hoga. b

Step-by-step explanation:

$cot(a + b)$

Answer 2

Answer:

We have given that ;

$\rm \frac{x}{y} = \frac{cosA}{cosB}$

$\large:\longmapsto \rm \underline{ \underline{x = y. \frac{cosA}{cosB} }} \small - - - - (1)$

Now we have to calculate :

$\: \: \: \rm\frac{x \: tanA + y \: tanB}{x + y}$

Substituting Value of x from (1) ;

$= \rm \dfrac{y \: \dfrac{cosA}{cosB} \times tan A + y \: \dfrac{sinB}{cosB} }{y \: \dfrac{cosA}{cosB} + y } \\ \\ :\implies \rm \dfrac{y \: \dfrac{ \cancel{cosA}}{cosB} \times \dfrac{sinA}{ \cancel{cosA}} + y \: \dfrac{sinB}{cosB} }{y \: \dfrac{cosA}{cosB} + y } \\ \\ = \rm \dfrac{y \: \dfrac{ {sinA}}{cosB} + y \: \dfrac{sinB}{cosB} }{y \: \dfrac{cosA}{cosB} + y } \\ \\ \rm = \dfrac{ \cancel y \bigg[ \dfrac{sinA}{cosB} + \dfrac{sinB}{cosB} \bigg]}{ \cancel y\bigg[ \dfrac{cosA}{cosB} + 1 \bigg]} \\ \\ \rm = \dfrac{ \dfrac{sinA}{cosB} + \dfrac{sinB}{cosB} }{ \dfrac{cosA}{cosB} + 1 } \\\rm = \frac{ \dfrac{sinA + sinB}{ \cancel{cosB}} }{\dfrac{cosA + cosB}{ \cancel{cosB}}} \\ \\ \rm = \frac{sinA + sinB}{cosA + cosB} \\ \\ \rm = \dfrac{ \cancel2 sin\bigg( \dfrac{A + B}{2} \bigg).\cancel{cos \bigg( \dfrac{A - B}{2} \bigg)}}{ \cancel2.cos \bigg( \dfrac{A + B}{2} \bigg). \cancel{cos \bigg( \dfrac{A - B}{2} \bigg)}} \\ \\ \rm = \dfrac{sin \bigg( \dfrac{A + B}{2} \bigg)}{cos \bigg( \dfrac{A + B}{2} \bigg)}$

$\large \bf\pink{ =tan \Bigg( \dfrac{A + B}{2} \Bigg)}$