Question

pls solve the above matching​

Answer 1

Answer:

option a) is correct.

C , D , B , A

Answer 2

Answer:

Given,

$\frac{sinx}{a} = \frac{cosx}{b} = \frac{tanx}{c}$

To Find :-

1) bc

2) a² + b²

3) $\sf\dfrac{1}{ck}$ + $\sf\dfrac{ak}{1 + bk}$

4)a² + b² + c²

$\huge\sf\underline{solution}$

$\sf\frac{sinx}{a}$ = k --- (1)

$\sf\frac{cosx}{b}$ = k ---(2)

$\sf\frac{tanx}{c}$ = k ----(3)

equation 2 × equation 3

$\frac{cosx \times tanx}{bc} = k \times k$

$\frac{cosx \times \frac{sinx}{cosx} }{bc} = k {}^{2}$

$\sf\therefore$ tanx = $\sf\frac{sinx}{cosx}$

$\frac{sinx}{bc} = k {}^{2}$

$bc = \frac{sinx}{ {k}^{2} }$

from equation --(1)

$sinx = ak$

$\sf\implies$ bc = $\sf\frac{ak}{k²}$

$\sf\therefore$ bc = $\sf\frac{a}{k}$

2) a² + b²

from equation (1):-

a = $\sf\frac{sinx}{k}$

"squaring on both sides"

a² = $\sf\frac{sin²x}{k²}$

from equation -- (2) :-

b = $\sf\dfrac{cosx}{k}$

"squaring on both sides"

b² = $\sf\dfrac{cos²x}{k²}$

${a}^{2} + {b}^{2} = \frac{sin {}^{2}x + {cos}^{2}x }{ {k}^{2} }$

${a}^{2} + {b}^{2} = \frac{1}{ {k}^{2} }$

3)$\sf\dfrac{1}{ck}$ + $\sf\dfrac{ak}{1 + bk}$

4) a² + b² + c²

Note :- 3,4 sums are in the attachment.