Question

solve this please prove this​

Answer 1

Given :-

$\rm \dfrac{Sin \: A}{1 + cos \: A} + \rm \dfrac{cos\: A}{sin \: A} = \dfrac{1}{\rm{Sin \: A} }$

To prove :-

LHS = RHS

Proof :-

$\rm \: R.H.S = \dfrac{1}{\rm{Sin \: A} }$

$\rm \: L.H.S = \rm \dfrac{Sin \: A}{1 + cos \: A} + \rm \dfrac{cos\: A}{sin \: A}$

$\rm \longrightarrow\dfrac{Sin \: A}{1 + cos \: A} + \rm \dfrac{cos\: A}{sin \: A} \\ \\ \\ \rm \longrightarrow\dfrac{ {Sin}^{2} \: A +cos \: A + {cos}^{2} \: A }{(1 + cos \: A)sin \: A} \\ \\ \\ \rm \longrightarrow\dfrac{ {Sin}^{2} \: A + {cos}^{2} \: A +cos \: A}{(1 + cos \: A)sin \: A} \\ \\ \\ \rm \longrightarrow\dfrac{ 1 +cos \: A}{(1 + cos \: A)sin \: A} \\ \ \\ \\ \rm \longrightarrow\dfrac{1}{sin \: A}$

Therefore, L.H.S = R.H.S

hence proved

Additional Information :-

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$

Answer 2

${\colorbox{cyan}{✿QuesTion}}$

TO PROVE:-

$\bold{ \frac{sin \:A }{1 + cos \:A} + \frac{cos \:A}{sin \:A } = \frac{1}{ sin \: A } }$

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${\colorbox{pink}{✿ AnsWer}}$

${\boxed{\underline{\red{ L.H.S}}}}$

$: \longrightarrow \: \frac{sin \: A }{1 + \: cos \: A } + \frac{cos \: A }{sin A }$

✮ TAKING L.C.M:-

L.C.M is (1+cos A ) sin A

$: \longrightarrow \: \frac{ {sin}^{2}\:A \: + cos \:A \: + {cos}^{2}A}{(1 + cos \: A)sin \:A }$

$: \longrightarrow \: \frac{ {sin}^{2} A \: + {cos}^{2}A \: + cos \: A }{(1 + cos \: A)sin \: A }$

AS,

${ \boxed{ \underline{ \purple{Sin^2 A + Cos^2 A + Cos A = 1+ Cos A}}}}$

$: \longrightarrow \: \frac{ \cancel{1 + cos \:A}}{ \:( \cancel{1 + cos \:A }) sin \: A }$

$: \longrightarrow \: \orange{\frac{1}{sin \: A} }$

${\boxed{\underline{\red{ R.H.S}}}}$

THEREFORE,

LHS = RHS

HENCE PROVED....

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★MORE TO KNOW:-

$⬤ \: sin \: (a + b) = \: sin \: a \: co</u><u>s</u><u> b \: + \: cos \: \: sin \: b \:$

$⬤ sin \: (a - b) = \: sin \: a \: cos \: b - cos \: a \: sin \: b \:.$

$⬤ cos(a + b) = cos \: a \: cos \: b \: - \: sin \: a \: sin \: b \: .$

$⬤ \: cos \: (a - b) = cos \: a \: cos \: b \: + sin \: a \: sin \: b.$

$⬤ tan(a + b) = \frac{tan \: a \: + \: tan \: b}{1 - tan \: a \: tan \: b \: }$

$⬤ tan(a - b) = \frac{tan \: a \: - \: tan \: b}{1 + \: tan \: a \: tan \: b \: }$