show that d/dx (1/3 tan³x) = tan²x sec²x
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Answer:
Correct option is
D
1
Given,
xsinθ=ycosθ
y=
cosθ
xsinθ
xsin
3
θ+ycos
3
θ=sinθcosθ
xsin
3
θ+
cosθ
xsinθ
×cos
3
θ=sinθcosθ
xsin
3
θ+xsinθcos
2
θ=sinθcosθ
xsinθ(sin
2
θ+cos
2
θ)=sinθcosθ
⇒xsinθ=sinθcosθ
∴x=cosθ
y=
cosθ
xsinθ
=
cosθ
cosθ×sinθ
∴y=sinθ
∴x
2
+y
2
=cos
2
θ+sin
2
θ=1
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