Math, asked by jaywantkhot30, 3 months ago

34) Prove that Cot A
& sec² A
sinA-1
(1+seca
ma
1+ sinA​

Answers

Answered by DeathAura
0

Answer:

LHS =

1+sinA

cot

2

A(secA−1)

=

sin

2

A(1+sinA)×cosA

cos

2

A(1−cosA)

=

sin

2

A(1+sinA)

cosA(1−cosA)

×

(1−sinA)(1+cosA)

(1−sinA)×(1+cosA)

=

(1+cosA)

(1−sinA)

×

cosA

1

=

cosA×cosA(1+secA)

1×(1−sinA)

=sec

2

A(

1+secA

1−sinA

)

=RHS

Answered by Anonymous
0

Answer:

here is ur answer dude

Step-by-step explanation:

cot²A(secA-1/1+sinA)+sec²A(sinA-1/1+secA)=0

L.H.S.

cot²A(secA-1/1+sinA)+sec²A(sinA-1/1+secA)

=cot²A[(secA-1)(1-sinA)/(1+sinA)(1-sinA)]+sec²A[(sinA-1)(secA-1)/(secA+1)(secA-1)]

=cot²A[(secA-secA.sinA-1+sinA)/(1-sin²A)]+sec²A[(sinA.secA-sinA-secA+1)/(sec²A-1)]

=cot²A[(secA-secA.sinA-1+sinA)/cos²A]+sec²A[(sinA.secA-sinA-secA+1)/tan²A]

=(cos²A/sin²A)[(secA-secA.sinA-1+sinA)/cos²A]+[sec²A(sinA.secA-sinA-secA+1)](cos²A/sin²A)

=[(secA-secA.sinA-1+sinA)/sin²A]+[(sinA.secA-sinA-secA+1)/sin²A]

=(secA-secA.sinA-1+sinA+sinA.secA-sinA-secA+1)/sin²A

=0/sin²A

=0

R.H.S=L.H.S

Hence proved…

PLEASE MY ANSWER AS BRAINLIST

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