sin theta minus cos theta by sin theta + cos theta + sin theta + cos theta whole divided by sin theta minus cos theta is equal to 2 by 2 sin squared theta minus one
Answers
Answer:
Step-by-step explanation:
given sinα -cosα /sinα +cosα +sin α +cosα/sin α -cos α =2/2sin²-1
L.H.S.
take lcm
⇒[( sinα -cosα)²+ ( sinα +cosα)²]/[( sinα -cosα)( sinα +cosα)]
⇒2(sin²α+cos²α) /(sin²α-cos²α) [ ∵ (a+b)²+(a-b)²= 2(a²+b²)]
⇒2/(cos 2α) [ ∵ sin²α-cos²α=cos 2α]
⇒2/ 2sin²α-1 [ ∵ cos 2α= 2sin²α-1 ]
L.H.S.=R.H.S.
sinθ−cosθ
sinθ+cosθ
+
sinθ+cosθ
sinθ−cosθ
=\frac{{(sin\theta+cos\theta)}^2+{(sin\theta-cos\theta)}^2}{(sin\theta-cos\theta)(sin\theta+cos\theta)}=
(sinθ−cosθ)(sinθ+cosθ)
(sinθ+cosθ)
2
+(sinθ−cosθ)
2
=\frac{{(sin^2\theta+cos^2\theta+2sin\theta.cos\theta)}+{(sin^2\theta+cos^2\theta-2sin\theta.cos\theta)}}{(sin^{2}\theta-cos^{2}\theta)}=
(sin
2
θ−cos
2
θ)
(sin
2
θ+cos
2
θ+2sinθ.cosθ)+(sin
2
θ+cos
2
θ−2sinθ.cosθ)
=\frac{{(1+2sin\theta.cos\theta)}+{(1-2sin\theta.cos\theta)}}{(sin^{2}\theta-cos^{2}\theta)}=
(sin
2
θ−cos
2
θ)
(1+2sinθ.cosθ)+(1−2sinθ.cosθ)
=\frac{2}{(sin^{2}\theta-cos^{2}\theta)}=
(sin
2
θ−cos
2
θ)
2
=\frac{2}{sin^{2}\theta-(1-sin^{2}\theta)}=
sin
2
θ−(1−sin
2
θ)
2
=\frac{2}{sin^{2}\theta-1+sin^{2}\theta}=
sin
2
θ−1+sin
2
θ
2
=\frac{2}{2sin^{2}\theta-1}=
2sin
2
θ−1
2