Math, asked by ANILAV6330, 3 months ago

(secA + cosecA)^2=(1+tanA)^2+(1+cotA)^2

Answers

Answered by Anonymous
2

Step-by-step explanation:

Hence, the values ofA satisfying the equation

(

1

+

tan

A

)

2

+

(

1

+

cot

A

)

2

=

(

sec

A

+

csc

A

)

2

are,

A

=

...

...

,

7

π

2

,

3

π

,

5

π

2

,

2

π

,

3

π

2

,

π

,

π

2

,

0

,

π

2

,

π

,

3

π

2

,

2

π

,

5

π

2

,

3

π

,

7

π

2

,

...

...

Explanation:

(

1

+

tan

A

)

2

+

(

1

+

cot

A

)

2

=

(

sec

A

+

csc

A

)

2

tan

A

=

sin

A

cos

A

cot

A

=

cos

A

sin

A

sec

A

=

1

cos

A

csc

A

=

1

sin

A

(

1

+

sin

A

cos

A

)

2

+

(

1

+

cos

A

sin

A

)

2

=

(

1

cos

A

+

1

sin

A

)

2

(

cos

A

+

sin

A

)

2

cos

2

A

+

(

sin

A

+

cos

A

)

2

sin

2

A

=

(

sin

A

+

cos

A

)

2

sin

A

cos

A

(

cos

A

+

sin

A

)

2

(

1

cos

2

A

+

1

sin

2

A

)

=

(

cos

A

+

sin

A

)

2

(

1

sin

A

cos

A

)

sin

2

A

+

cos

2

A

sin

2

A

cos

2

A

=

1

sin

A

cos

A

1

(

sin

A

cos

A

)

2

=

1

sin

A

cos

A

(

sin

A

cos

A

)

2

=

sin

A

cos

A

(

sin

A

cos

A

)

2

sin

A

cos

A

=

0

sin

A

cos

A

(

sin

A

cos

A

1

)

=

0

sin

A

=

0

,

cos

A

=

0

,

sin

A

cos

A

1

=

0

sin

A

=

0

A

=

0

,

π

,

2

π

,

3

π

,

...

cos

A

=

0

A

=

π

2

,

3

π

2

,

5

π

2

,

7

π

2

,

...

...

sin

A

cos

A

=

0

1

2

sin

2

A

=

0

sin

2

A

=

0

2

A

=

0

,

π

,

2

π

,

3

π

,

...

A

=

0

,

π

2

,

π

,

3

π

2

,

2

π

,

5

π

2

,

3

π

,

7

π

2

,

...

...

Hence, the values ofA satisfying the equation

(

1

+

tan

A

)

2

+

(

1

+

cot

A

)

2

=

(

sec

A

+

csc

A

)

2

are,

A

=

...

...

,

7

π

2

,

3

π

,

5

π

2

,

2

π

,

3

π

2

,

π

,

π

2

,

0

,

π

2

,

π

,

3

π

2

,

2

π

,

5

π

2

,

3

π

,

7

π

2

,

...

...

Answered by itsme12341
0

Step-by-step explanation:

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