Math, asked by mekadex962, 6 months ago

prove that cosA / 1 - sinA = 2secA - cosA/ 1+sinA

Answers

Answered by anishdurgam85
1

Answer:

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Step-by-step explanation:

\frac{cos A}{1} = cosA

cos A - sin A = 2 sec - \frac{cosA}{1+sin A}

Bring \frac{cosA}{1+sin A} to LHS

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

={cos²A +(1+sinA)²}/cosA.(1+sinA)

={cos²A+1+sin²A+2sinA}/cosA.(1+sinA)

use sin²∅ +cos²∅ =1

=(1+1+2sinA)/cosA(1+sinA)

=2(1+sinA)/cosA(1+sinA)

=2/cosA

=2secA = RHS

Hope this helps you!

Answered by ItzArchimedes
4

Solution :-

• cosA/ [ 1 - sinA ] = 2secA - cosA / [ 1 + sinA ]

By simplifying ,

• cosA / [ 1 - sinA ] + cosA / [ 1 + sinA ] = 2 secA

That means we need to prove cosA / [ 1 - sinA ] + cosA / [ 1 + sinA ] = 2 secA

Taking LHS & simplifying ,

⇒ cosA [ 1 + sinA ] + cosA [ 1 - sinA ] / [ 1 + sinA ] [ 1 - sinA ]

⇒ cosA + sinAcosA + cosA - cosAsinA / 1² - sin²A

[ ( a + b ) ( a - b ) = - ]

⇒ 2 cosA/cos²A

[ 1 - sin²A = cos²A ]

⇒ 2 × cosA/cos²A

⇒ 2 × [ 1 / cosA ]

⇒ 2 × secA

[ 1 / cosA = secA ]

2secA

Now , comparing with RHS

2secA = 2secA

LHS = RHS

Hence , proved

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