Question

prove that sin square alpha by cot alpha + cos square alpha by tan alpha + 2 sin alpha cos alpha is =sec alpha cosec alpha =tan alpha +cot alha​

Answer 1

Explanation:

Since we have given that

(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)=\sec \alpha+\cosec \alpha(sinα+cosα)(tanα+cotα)=secα+cosecα

We need to prove it .

From L.H.S. ,

\begin{lgathered}(\sin \alpha+\cos \alpha)(\tan \alpha+\cot \alpha)\\\\=(\sin \alpha+\cos \alpha)(\frac{\sin \alpha}{\cos \alpha}+\frac{\cos\alpha}{\sin \alpha})\\\\=(\sin \alpha+\cos \alpha)(\frac{\sin^2\alpha+\cos^2\alpha}{\sin \alpha \cos \alpha})\\\\=(\sin \alpha+\cos \alpha)(\frac{1}{\sin \alpha \cos \alpha})\\\\=(\frac{\sin \alpha}{\sin \alpha \cos \alpha})+(\frac{\cos \alpha}{\sin \alpha \cos \alpha})\\\\=\sec \alpha+cosec\ \alpha\end{lgathered}

(sinα+cosα)(tanα+cotα)

=(sinα+cosα)(

cosα

sinα

+

sinα

cosα

)

=(sinα+cosα)(

sinαcosα

sin

2

α+cos

2

α

)

=(sinα+cosα)(

sinαcosα

1

)

=(

sinαcosα

sinα

)+(

sinαcosα

cosα

)

=secα+cosec α

Answer 2

Aɴꜱᴡᴇʀ

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Gɪᴠᴇɴ

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ᴛᴏ ᴘʀᴏᴠᴇ

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Sᴛᴇᴘꜱ

First we take LHS

$\begin{lgathered}= (\sin \alpha + cos \alpha)(tan \alpha + cot \alpha) \\ \\ = ( \sin \alpha + \cos \alpha )( \: \: \frac{ \sin \alpha }{ \: \: \: \cos \alpha } + \frac{ \cos \alpha }{ \sin \alpha } \: \: )\end{lgathered}$

$= [sin (alpha) + cos (alpha)] [ ({sin}^2{alpha}+{cos}^{alpha})/cos alpha × sin alpha ]$

= [ sin alpha + cos alpha] [ 1/ (cos alpha ×sin alpha )]

= [ sin alpha + cos alpha] / [ sin alpha × cos alpha]

Now we take RHS

= sec alpha + cosec alpha

= 1/cos alpha + 1/sin alpha

= [ sin alpha + cos alpha] / [sin alpha × cos alpha ]

Now ,

LHS = RHS

HENCE PROVED

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$\huge{\mathfrak{\purple{hope\; it \;helps}}}$