Question

please help me
what is the next step
i don't known
please solve the problem
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Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Prove :- cosA/(1 - tanA) + sinA/(1-cotA) = cosA + sinA

Sᴏʟᴜᴛɪᴏɴ :-

Solving LHS, (Like You did by using tanA = (sinA/cosA) & cotA = (cosA/sinA) in denominator , we get,

→ cos A/(1 - sin A/cos A) + sin A/(1 - cos A/sin A)

→ cosA /(cosA - sinA)/cosA + sinA/(sinA - cosA) / sinA

Now, using a/(b/c) = a * (c/b) , we get, cosA & sinA will come in Numerator, so we get ,

→ cos²A/ (cosA - sinA) + sin²A / (sinA - cosA)

Now, taking (-1) common from second part Denominator , we get,

→ cos²A/ (cos A - sin A) - sin²A / (cosA - sinA)

Now, taking LCM ,

→ (cos²A - sin²A) / (cos A - sin A)

using (a² - b²) = (a + b)(a - b) in Numerator now,

→ (cosA - sinA)(cosA + sinA) / (cosA - sinA)

(cosA - sinA) will be cancel now,

→ cos A + sin A = RHS = Proved.

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

$\star {\sf{ \: \frac{cosA}{ 1 - tan A} + \frac{sinA}{1 - cotA} }}$

${\bf{\blue{\underline{Now:}}}}$

We know that,

$: {\star{\boxed {\sf{ \: tanA = \frac{sinA}{cosA} }} \: \: \: \: \star{\boxed{ \sf{ \: cotA = \frac{cosA}{sinA} }}}}}$

$: \implies{\sf{ \frac{cosA}{1 - \frac{sinA}{cosA} } + \frac{sinA}{1 - \frac{cosA}{sinA} } }}$

$: \implies{\sf{ \frac{cosA}{ \frac{cosA - sinA}{cosA} } + \frac{sinA}{ \frac{sinA - cosA}{sinA} } }}$

$: \implies{\sf{ \frac{cos ^{2} A }{ {cosA - sinA} } + \frac{sin ^{2} A }{ {sinA - cosA}} }} \\ \\ </p><p></p><p>$

$: \implies{\sf{ \frac{cos ^{2} A }{ {cosA - sinA} } - \frac{sin ^{2} A }{ {cosA - sinA}} }}$

$: \implies{\sf{ \frac{cos ^{2} A - sin ^{2} A }{ {cosA - sinA} } }}$

${\star{\boxed {\sf{\purple{ \: ( {x}^{2} - {y}^{2} )= (x + y)(x - y) }}}}}$

$: \implies{\sf{ \frac{(cos A - sin A )(cos A + sin A )}{ {cosA - sinA} } }}$

$: \implies{\sf{ cos A + sin A }}$

Hence proved!