Question

If sinθ+cosθ=a and tanθ+cotθ=b , then b(a2−1)=​

Answer 1

Answer:

...............................

Answer 2

Answer:

$\huge{ \bf{ \underline{ \red{Given}}}}$

Sinθ+cosθ=a, tanθ+cotθ=b

Find:- b(a²-1)

Solution:-

Sinθ+Cosθ = a ........ (1)

Tanθ+Cotθ =b .......... (2)

From equation (2)...

${ \boxed{ \sf{tanθ = \frac{sinθ}{cosθ} }}}$

${ \boxed{ \sf{cotθ= \frac{cosθ}{sinθ} }}}$

Tanθ+cotθ= b

${ \sf\frac{sinθ}{cosθ} + \frac{cosθ}{sinθ} = b }$

${ \sf \frac{ {sinθ}^{2} + {cosθ}^{2} }{sinθcosθ} = b}$

${ \boxed{ \therefore{ \sf{ {sinθ}^{2} + {cosθ}^{2} = 1 }}}}$

${ \sf \frac{1}{sinθcosθ} = b}$

${ \sf{sinθcos θ= \frac{1}{b} ........(3)}}$

Take equation (1) and do squaring on both sides.

${ \sf{( {sinθ + cosθ) }^{2} = {a}^{2} }}$

${ \sf{sinθ}^{2} + {cosθ}^{2} + 2sinθcos θ= {a}^{2} }.......(4)$

We know that

${ \boxed{ \sf{ {sinθ}^{2} + {cosθ}^{2} = 1 }}}$

And, from equation (3)

${ \boxed{ \sf{sinθcosθ = \frac{1}{b} }}}$

Substitute these values in equation (4)

${ \sf{1 + 2( \frac{1}{b} ) = {a}^{2} }}$

${ \sf{2( \frac{1}{b} ) = {a}^{2} - 1}}$

${ \sf{ \frac{2}{b} = {a}^{2} - 1}}$

${ \sf{2 = b( {a}^{2} - 1)}}$

${ \therefore{ \sf{b( {a}^{2} - 1) = 2}}}$