Question

Prove that -2sinacosa= -2/tana+cota

Answer 1

To prove--->

-2SinA CosA = -2 / ( tanA + CotA )

Proof---> First we find value of ( tanA + CotA )

We know that ,

tanA = SinA / CosA , CotA = CosA / SinA

tanA + CotA = ( SinA / CosA ) + ( CosA / SinA )

= ( Sin²A + Cos²A ) / SinA CosA

We know that Sin²A + Cos²Α = 1 , applying it , we get,

= 1 / SinA CosA

Now taking ,

RHS = -2 / ( tanA + CotA )

= -2 / ( 1 / SinA CosA )

= -2 SinA CosA = LHS

Additional information--->

1) Sin²θ + Cos²θ = 1

2) 1 + tan²θ = Sec²θ

3) 1 + Cot²θ = Cosec²θ

4) Sin( 90° - θ ) = Cosθ

5) Cos( 90° - θ ) = Sinθ

6) tan(90° - θ ) = Cotθ

#Answerwithquality&BAL

Answer 2

RHS

$\frac{ - 2}{tana + cota} \\ \\ = \frac{ - 2}{ \frac{sina}{cosa} + \frac{cosa}{sina} } \\ \\ = \frac{ - 2}{ \frac{ {sin}^{2} a + {cos}^{2}a }{sinacosa} } \\ \\ = \frac{ - 2}{ \frac{1}{sinacosa} } \\ \\ = - 2sinacosa$

LHS