Question

prove that cosecA-sinA secA-cosA =1/tanA+cotA
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Answer 1

Question :

$\bf{(cosecA - sinA)(secA - cosA) = \dfrac{1}{tanA + cotA}}$

Solution :

Here ,

• LHS = $\bf{(cosecA - sinA)(secA - cosA)}$

• RHS = $\bf{\dfrac{1}{tanA + cotA}}$

By solving the RHS , we get :

$:\implies \bf{\dfrac{1}{tanA + cotA}}$

We know that :-

• $\bf{tan\theta = \dfrac{sin\theta}{cos\theta}}$

• $\bf{cot\theta = \dfrac{cos\theta}{sin\theta}}$

By Substituting the values in tanA and cotA , we get :

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

$:\implies \bf{\dfrac{1}{\dfrac{sin^{2}A + cos^{2}A}{cosAsinA}}}$

$:\implies \bf{\dfrac{1}{sin^{2}A + cos^{2}A} \times cosAsinA}$

$:\implies \bf{\dfrac{cosAsinA}{sin^{2}A + cos^{2}A}}$

We know that :-

$\bf{sin^{2}A + cos^{2}A = 1}$

By Substituting the values in it, we get :

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

$:\implies \bf{cosAsinA}$

$\boxed{\therefore \bf{RHS = cosAsinA}}$

Hence RHS is cosAsinA.

By solving the LHS, we get :

$:\implies \bf{(cosecA - sinA)(secA - cosA)}$

We know that :-

• $\bf{cosec\theta = \dfrac{1}{sin\theta}}$

• $\bf{sec\theta = \dfrac{1}{cos\theta}}$

By Substituting the value of cosecA and secA , we get :

$:\implies \bf{\bigg(\dfrac{1}{sinA} - sinA\bigg)\bigg(\dfrac{1}{cosA} - cosA\bigg)}$

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

We know that ;

• $\bf{cos^{2}A = 1 - sin^{2}A}$

• $\bf{sin{2}A = 1 - cos^{2}A}$

By Substituting them in the equation , we get :

$:\implies \bf{\dfrac{cos^{2}A}{sinA} \times \dfrac{sin^{2}A}{cosA}}$

$:\implies \bf{\dfrac{cos^{2}A}{cosA} \times \dfrac{sin^{2}A}{sinA}}$

$:\implies \bf{cosAsinA}$

$\boxed{\therefore \bf{RHS = cosAsinA}}$

Hence LHS is cosAsinA.

By putting the LHS and RHS together , we get :

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

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

$\boxed{\therefore \bf{(cosecA - sinA)(secA - cosA) = \dfrac{1}{tanA + cotA}}}$

Hence we get that LHS = RHS.

Proved !!

Answer 2

Step-by-step explanation:

$\pink{ \huge \mathfrak {Solution}}$

L.H.S

$\sf \bigg(cosesA - sinA \bigg) \bigg(secA - cosA \bigg) \\ \\ \\ \sf \implies\bigg( \frac{1}{sinA} - sinA \bigg) \bigg( \frac{1}{cosA} - cosA \bigg) \\ \\ \\ \sf \implies \bigg( \frac{1 - {sin}^{2}A }{sinA} \bigg) \bigg( \frac{1 - {cos}^{2}A }{cosA} \bigg) = \frac{ {cos}^{ \cancel2} A \: \: {sin}^{ \cancel2}A} { \cancel{sinA \: \: cosA} } \\ \\ \\ \sf \: \frac{1}{tanA + cotA} =R.H.S \\ \\ \\ \sf \: L.H.S = R.H.S$