Math, asked by jlsea2019, 1 year ago


((cosec A - cot A)2 +1)/
(sec A (cosec A - cot A))
= 2 cot A

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Answers

Answered by RvChaudharY50
34

Qᴜᴇsᴛɪᴏɴ :-

Prove :- {(cosecA - cotA)² +1}/(secA(cosecA - cotA) = 2cotA

Sᴏʟᴜᴛɪᴏɴ :-

Taking LHS,

→ {(cosecA - cotA)² +1}/(secA(cosecA - cotA)

Putting 1 = cosec²A - cot²A in Numerator,

→ {(cosecA - cotA)² +(cosec²A - cot²A)}/(secA(cosecA - cotA)

using a² - b² = (a + b)(a - b) Now,

→ {(cosecA - cotA)² +(cosecA - cotA)(cosecA + cotA)}/(secA(cosecA - cotA)

Taking (cosecA - cotA) common from Numerator,

→ (cosecA - cotA){(cosecA - cotA) + (cosecA + cotA)} / secA(cosecA - cotA)

→ {(cosecA - cotA) + (cosecA + cotA)} / secA

→ 2cosecA / secA

Putting cosecA = (1/sinA) & secA = (1/cosA) Now,

→ 2*(1/sinA) / (1/cosA)

→ (2/sinA) * (cosA/1)

→ (2cosA/sinA)

2cotA = RHS (Proved).

Answered by Anonymous
9

QUESTION:

 \frac{( { \csc \: a -  \cot \: a)  }^{2}  + 1}{ \sec \: a ( \csc \: a -  \cot \: a \: )  }  = 2 \cot(a)

ANSWER:

We use the trigonometry identity here;

 { \csc( \alpha ) }^{2}  -  { \cot( \alpha ) }^{2}  = 1 \\  \\  \csc( \alpha )  =  \frac{1}{ \sin( \alpha ) }  \\  \\  \sec( \alpha )  =  \frac{1}{ \cos( \alpha ) }  \\  \\  \\  \cot( \alpha )  =  \frac{ \cos( \alpha ) }{ \sin( \alpha ) }

other identity we used here :

 {x}^{2}   -  {y}^{2}  = (x + y)(x - y)

NOW COME TO YOUR QUESTION :

TAKING LHS;

I LET A AS ALPHA

 \frac{( { \csc( \alpha ) -  \cot( \alpha )  }^{2}  + 1 }{ \sec( \alpha ) ( \csc( \alpha )  -  \cot( \alpha ) }

using the identity

 \frac{( { \csc( \alpha )  -  \cot( \alpha ) }^{2}  + ( { \csc( \alpha ) }^{2}  -  { \cot( \alpha ) }^{2} }{ \sec( \alpha ) ( \csc( \alpha )  -  \cot( \alpha ) }

 \frac{( { \csc( \alpha ) -  \cot( \alpha )  }^{2} + ( \csc( \alpha )  -  \cot( \alpha ) ( \csc( \alpha ) +  \cot( \alpha )   }{ \sec( \alpha ) ( \csc( \alpha ) -  \cot( \alpha )  }

TAKING common here

 \frac{( \csc( \alpha ) -  \cot( \alpha )   + ( \csc( \alpha )  +  \cot( \alpha ) }{ \sec( \alpha ) }

 \frac{2 \csc( \alpha ) }{ \sec( \alpha ) }

 \frac{2 \times  \frac{1}{ \sin( \alpha ) } }{ \frac{1}{ \cos( \alpha ) } }  \\ 2 \times  \frac{  \cos( \alpha ) }{ \sin( \alpha ) }

2 \cot( \alpha )

LHS = RHS

HENCE PROVED

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