Question

Prove that :-

(coses A - sin A) ( sec A - cos A) = 1/tanA+cotA

Answer 1

Answer:

We have to just simplify both side of the question.

Now, taking L. H. S =

(cosecA - sinA)×(secA - cosA)

= (1/sinA - sinA)×(1/cosA - cosA)

= (1-sin²A)/sinA × (1-cos²A)/cosA

= cos²A/sinA × sin²A/cosA

= cos²Asin²A/cosAsinA

= cosAsinA

Now, R. H. S =

1/(tanA + cotA)

= 1/(sinA/cosA + cosA/sinA)

= 1/[(sin²A + cos²A)/cosAsinA]

= 1/[(1/cosAsinA)]

= cosAsinA

Since L. H. S. = R. H. S.

Therefore the given equation is proof.

In such types of question you have to take care of braket and sign.

Thanks!

Answer 2

$\sf{ \underline{TO \: PROVE :-}}$

$\displaystyle{\implies{ \sf{(coses \:A - sin \: A)(sec \: A - cos \: A) = \frac{1}{tan \: A + cot \: A} }}}$

$\sf{ \underline{SOLUTION :-}}$

• Let's take L.H.S. –

$\displaystyle{\implies{ \sf{(coses \:A - sin \: A)(sec \: A - cos \: A)}}}$

$\displaystyle{\implies\sf\:\left(\:\:\dfrac{ 1 }{sin \: A } - \frac{sin \: A}{1} \:\right)} \sf \:\left(\:\:\dfrac{ 1 }{cos \: A } - \frac{cos \: A}{1} \:\right) \: \: \: \: \\ \sf Since,\: cosec \: A = \frac{1}{sin \:A } \: and \: sec \: A \: = \frac{1}{cos \:A }$

$\displaystyle{\implies\sf\:\left(\:\:\dfrac{ 1 - sin^{2} A }{sin \: A } \:\right)} \sf \:\left(\:\:\dfrac{ 1 - cos ^{2} A }{cos \: A } \:\right)$

$\displaystyle{\implies\sf\:\left(\:\:\dfrac{ cos ^{ \cancel2}A \: \: sin^{ \cancel2}A }{ \cancel{sinA} \: \cancel{ cosA} } \:\right)}$

$\displaystyle \implies{ \sf{sin \: A \: cos \: A = L.H.S.}}$

• Now, Let's take R.H.S –

$\displaystyle{ \implies{ \sf{\frac{1}{tan \: A + cot \: A}}}}$

$\displaystyle {\implies \sf{ \frac{1}{ \frac{sinA}{cosA} + \frac{cosA}{sinA} }}}$

$\displaystyle {\implies \sf{ \frac{1}{ \frac{sin^{2} A \: + \: cos ^{2} A}{cos A \: sinA} }}}$

$\displaystyle {\implies \sf{ \frac{ \cancel{sinA} \: \: \cancel{cosA}}{sin ^{ \cancel2} A + cos ^{ \cancel2}A } }}$

$\displaystyle \implies{ \sf{sin \: A \: cos \: A = R.H.S.}}$

$\displaystyle \implies \sf{L.H.S = R.H.S}$

$\displaystyle{ \implies{ \sf{\bold{Hence \:Proved}}}}$