Math, asked by Oshin18, 9 months ago

(cosec A -secA )(cot A-Tan A)=(cosec A+secA )(SecA cosecA -2 )

Answers

Answered by sahatapan1234567
1

Answer:

No answer to me. Very sorry.

Answered by sonujeet613
1

Answer:

To prove:

\displaystyle\frac{tanA+secA}{cosecA+cotA}+\frac{tanA-secA}{cosecA-cotA}=2(tanA\;cosecA-cotA\;secA)

cosecA+cotA

tanA+secA

+

cosecA−cotA

tanA−secA

=2(tanAcosecA−cotAsecA)

\text{Now,}Now,

\displaystyle\frac{tanA+secA}{cosecA+cotA}+\frac{tanA-secA}{cosecA-cotA}

cosecA+cotA

tanA+secA

+

cosecA−cotA

tanA−secA

=\displaystyle\frac{(tanA+secA)(cosecA-cotA)+(tanA-secA)(cosecA+cotA)}{(cosecA+cotA)(cosecA-cotA)}=

(cosecA+cotA)(cosecA−cotA)

(tanA+secA)(cosecA−cotA)+(tanA−secA)(cosecA+cotA)

=\displaystyle\frac{(tanA+secA)(cosecA-cotA)+(tanA-secA)(cosecA+cotA)}{cosec^2A-cot^2A}=

cosec

2

A−cot

2

A

(tanA+secA)(cosecA−cotA)+(tanA−secA)(cosecA+cotA)

\text{Using,}\bf\;cosec^2A-cot^2A=1Using,cosec

2

A−cot

2

A=1

=\displaystyle\;tanA\;cosecA-secA\;cotA+tanA\;cosecA-secA\;cotA=tanAcosecA−secAcotA+tanAcosecA−secAcotA

=\displaystyle\;2\;tanA\;cosecA-2\;secA\;cotA=2tanAcosecA−2secAcotA

=\displaystyle\;2(tanA\;cosecA-secA\;cotA)=2(tanAcosecA−secAcotA)

\therefore\boxed{\bf\frac{tanA+secA}{cosecA+cotA}+\frac{tanA-secA}{cosecA-cotA}=2(tanA\;cosecA-cotA\;secA)}∴

cosecA+cotA

tanA+secA

+

cosecA−cotA

tanA−secA

=2(tanAcosecA−cotAsecA)

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