Question

HEY GUYS !!!
please solve it..

Answer 1

Instead of α, I used A in solving this problem,

Answer:

Step-by-step explanation:

Given :

To find the value of k if,

$1-\frac{Sin^2A}{1+CotA} - \frac{Cos^2A}{1+TanA} = KSinACosA$

Solution :

By solving LHS,

We get,

LHS = $1-\frac{Sin^2A}{1+CotA} - \frac{Cos^2A}{1+TanA}$

$= 1-\frac{Sin^2A}{1+\frac{CosA}{SinA} } - \frac{Cos^2A}{1+\frac{SinA}{CosA} }$

$= 1-\frac{Sin^2A}{(\frac{SinA +CosA}{SinA}) } - \frac{Cos^2A}{(\frac{SinA+CosA}{CosA}) }$

$= 1-\frac{Sin^3A}{SinA +CosA} - \frac{Cos^3A}{{SinA+CosA} }$

$= 1-(\frac{Sin^3A}{SinA +CosA} + \frac{Cos^3A}{{SinA+CosA} })$

$= 1-(\frac{Sin^3A+Cos^3A}{SinA +CosA} )$

We know the identity that,

a³ + b³ = (a + b) (a² - ab + b²),

Here a = Sin A , b = Cos A

$1-\frac{(SinA + CosA)(Sin^2A - SinACosA + Cos^2A)}{SinA + CosA}$

$1 - (sin^2A + Cos^2A - Sin A Cos A)$

We know that,

Sin²A + Cos²A = 1

Hence,

$1 - (1 - SinACosA) = 1 - 1 + SinACosA = SinACosA$

Hence,

Sin A Cos A = K Sin A Cos A,

The value of ⇒ k = 1,.