Question

prove that 1- sinAcosA /cosA ( secA - cosecA ) × sin^2A - cos^2A / sin^3 A+ cos^3 A = sinA​

Answer 1

Answer:

Step-by-step explanation:

=1-sinAcosA/cosA(secA-cosecA)*sin^2A-cos^2A/sin^3A+cos^3A=sinA

=1-sinA.cosA/cosA(1/cosA-1/sinA) *(sinA-cosA)(sinA+cosA)/(sinA+cosA)(sin^2A+cos^2A-sinA.cosA)

= 1-sinA.cosA/cosA.sinA-cosA/sinA.cosA*(sinA-cosA)(sinA+cosA)/(sinA+cosA)(1-sinA.cosA)

= 1/1/sinA= sinA

Answer 2

Question :

Prove that :

$\bf{\dfrac{1 - sinAcosA}{cosA(secA - cosecA)} \times \dfrac{sin^{2}A - cos^{2}A}{sin^{3}A + cos^{3}A} = sinA}$

To find :

To prove that LHS = RHS.

Solution :

By solving the LHS , we get :

$:\implies \bf{\dfrac{1 - sinAcosA}{cosA(secA - cosecA)} \times \dfrac{sin^{2}A - cos^{2}A}{sin^{3}A + cos^{3}A}}$

We know that , $\bf{sec\:\theta = \dfrac{1}{cos\:\theta}}$ and $\bf{cosec\:\theta = \dfrac{1}{sin\:\theta}}$ , so using this and Substituting it in the equation , we get :

$:\implies \bf{\dfrac{1 - sinAcosA}{cosA\bigg(\dfrac{1}{cosA} - \dfrac{1}{sinA}\bigg)} \times \dfrac{sin^{2}A - cos^{2}A}{sin^{3}A + cos^{3}A}}$

$:\implies \bf{\dfrac{1 - sinAcosA}{cosA\bigg(\dfrac{sinA - cosA}{cosAsinA}\bigg)} \times \dfrac{sin^{2}A - cos^{2}A}{sin^{3}A + cos^{3}A}}$

Now using this identity and substituting it in the equation , we get :

$\bf{(a^{2} - b^{2} = (a - b)(a + b)}$

$:\implies \bf{\dfrac{1 - sinAcosA}{cosA\bigg(\dfrac{sinA - cosA}{cosAsinA}\bigg)} \times \dfrac{(sinA - cosA)(sinA + cos A)}{sin^{3}A + cos^{3}A}}$

Now using this identity and substituting it in the equation , we get :

$\bf{(a^{3} - b^{3} = (a + b)(a^{2} + b^{2} - ab)}$

$:\implies \bf{\dfrac{1 - sinAcosA}{cosA\bigg(\dfrac{sinA - cosA}{cosAsinA}\bigg)} \times \dfrac{(sinA - cosA)(sinA + cos A)}{(sinA + cosA)(sin^{2}A + cos^{2} - sinAcosA)}}$

We know that , $\bf{sin^{2}\theta + cos^{2}\theta = 1}$ , so by substituting it in the equation , we get :

$:\implies \bf{\dfrac{1 - sinAcosA}{cosA\bigg(\dfrac{sinA - cosA}{cosAsinA}\bigg)} \times \dfrac{(sinA - cosA)(sinA + cos A)}{(sinA + cosA)(1 - sinAcosA)}}$

By cancelling the like terms , we get :

$:\implies \bf{\dfrac{1}{\dfrac{1}{sinA}}}$

$:\implies \bf{sinA}$

$\boxed{\therefore \bf{LHS = sinA}}$

Now putting LHS and RHS together , we get :

$:\implies \bf{LHS = RHS}$

$:\implies \bf{sinA = sinA}$

Hence , LHS = RHS proved.

$\boxed{\therefore \bf{\dfrac{1 - sinAcosA}{cosA(secA - cosecA)} \times \dfrac{sin^{2}A - cos^{2}A}{sin^{3}A + cos^{3}A} = sinA}}$

Proved !!