Math, asked by kushalrm24, 2 months ago

pls answer it fast plssssssss​

Attachments:

Answers

Answered by ExElegant
17

Step-by-step explanation:

\large \underline \bold{Solution}:-

\rm{\dfrac{tan\theta - Sin\theta}{Sin^{3} \theta}}

\rm{\dfrac{tan\theta}{Sin^{3} \theta} - \dfrac{Sin\theta}{Sin^{3} \theta}}

\rm{\dfrac{Sin\theta}{Cos\theta Sin^{3} \theta} - \dfrac{1}{Sin^{2} \theta}}

\rm{\dfrac{Cosec^{2}\theta}{Cos\theta} - Cosec^{2}\theta}

\rm{Cosec^{2}\theta \bigg(\dfrac{1}{Cos\theta} - 1\bigg)}

\rm{Cosec^{2}\theta \bigg(\dfrac{1 - Cos\theta}{Cos\theta} \bigg)}

\rm{Cosec^{2}\theta \bigg(\dfrac{(1 - Cos\theta)(1 + Cos\theta)}{Cos\theta(1 + Cos\theta)}\bigg)}

\rm{Cosec^{2}\theta \bigg(\dfrac{(1 - Cos^{2}\theta)}{Cos\theta(1 + Cos\theta)} \bigg)}

\rm{Cosec^{2}\theta \bigg(\dfrac{Sin^{2}\theta}{Cos\theta(1 + Cos\theta)}\bigg)}

\rm{\dfrac{1}{\cancel{Sin^{2}\theta}} \bigg(\dfrac{\cancel{Sin^{2}\theta}}{Cos\theta(1 + Cos\theta)}\bigg)}

\rm{\bigg(\dfrac{1}{Cos\theta(1 + Cos\theta)}\bigg)}

\rm{\bigg(\dfrac{Sec\theta}{(1 + Cos\theta)} \bigg)}

\large \underline \bold{HENCE \: PROVE}

Answered by santhipriya01
1

Answer:

Sin3θtanθ−Sinθ

\rm{\dfrac{tan\theta}{Sin^{3} \theta} - \dfrac{Sin\theta}{Sin^{3} \theta}}Sin3θtanθ−Sin3θSinθ

\rm{\dfrac{Sin\theta}{Cos\theta Sin^{3} \theta} - \dfrac{1}{Sin^{2} \theta}}CosθSin3θSinθ−Sin2θ1

\rm{\dfrac{Cosec^{2}\theta}{Cos\theta} - Cosec^{2}\theta}CosθCosec2θ−Cosec2θ

\rm{Cosec^{2}\theta \bigg(\dfrac{1}{Cos\theta} - 1\bigg)}Cosec2θ(Cosθ1−1)

\rm{Cosec^{2}\theta \bigg(\dfrac{1 - Cos\theta}{Cos\theta} \bigg)}Cosec2θ(Cosθ1−Cosθ)

\rm{Cosec^{2}\theta \bigg(\dfrac{(1 - Cos\theta)(1 + Cos\theta)}{Cos\theta(1 + Cos\theta)}\bigg)}Cosec2θ(Cosθ(1+Cosθ)(1−Cosθ)(1+Cosθ))

\rm{Cosec^{2}\theta \bigg(\dfrac{(1 - Cos^{2}\theta)}{Cos\theta(1 + Cos\theta)} \bigg)}Cosec2θ(Cosθ(1+Cosθ)(1−Cos2θ))

\rm{Cosec^{2}\theta \bigg(\dfrac{Sin^{2}\theta}{Cos\theta(1 + Cos\theta)}\bigg)}Cosec2θ(Cosθ(1+Cosθ)Sin2θ)

\rm{\dfrac{1}{\cancel{Sin^{2}\theta}} \bigg(\dfrac{\cancel{Sin^{2}\theta}}{Cos\theta(1 + Cos\theta)}\bigg)}Sin2θ1(Cosθ(1+Cosθ)Sin2θ)

\rm{\bigg(\dfrac{1}{Cos\theta(1 + Cos\theta)}\bigg)}(Cosθ(1+Cosθ)1)

\rm{\bigg(\dfrac{Sec\theta}{(1 + Cos\theta)} \bigg)}((1+Cosθ)Secθ)

\large \underline \bold{HENCE \: PROVE}HENCEPROVE

please mark me as brainliest

Similar questions