(1 + cot theta + tan theta )(sin theta - cos theta) is equal to secant theta upon cosec squared theta minus cosec squared theta upon secant squared theta Prove it
Answers
Answered by
0
Step-by-step explanation:
Answer
LHS=
sec
3
θ−cosec
3
θ
(1+cotθ+tanθ)(sinθ−cosθ)
=
cos
3
θ
1
−
sin
3
θ
1
(1+
sinθ
cosθ
+
cosθ
sinθ
)(sinθ−cosθ)
(sin
2
θ+cos
2
θ=1)
=
sin
3
θ.cos
3
θ
sin
3
θ−cos
3
θ
(1+
cosθ.sinθ
cos
2
θ+sin
2
θ
)(sinθ−cosθ)
=
(sin
3
θ−cos
3
θ).sinθ.cosθ
(cosθ.sinθ+1)(sinθ−cosθ)
.sin
3
θ.cos
3
θ
We know that,
a
3
−b
3
=(a−b)(a
2
+ab+b
2
)
∴
(sinθ−cosθ)(sin
2
θ+sinθ.cosθ+cos
2
θ)
(cosθ.sinθ+1)(sinθ−cosθ)×sin
2
θ.cos
2
θ
=
(sinθ−cosθ)(1+sinθ.cosθ)
(cosθ.sinθ+1)(sinθ−cosθ)
.sin
2
θ.cos
2
θ
=sin
2
θ.cos
2
θ
=RHS
Similar questions