Math, asked by raimasaha29, 9 months ago

Prove that
sin (90° - A)/
I-tanA
+
sin A/
1-tan (90 - A)
= cosA+cos (90-A)​

Answers

Answered by unicorn276
4

Step-by-step explanation:

Given

sinA+cosA=

2

sin(90−A)

2

1

sinA+

2

1

cosA=sin(90−A)

⇒sinA.cos45

+cosA+sin45

=sin(90−A)

⇒(A+45

)=sin(90

−A)

Equating

A+45=90−A

⇒2A=45

⇒A=22

2

1

A=

8

π

∴cotA=cot

8

π

=

2

+1

Answered by seshathrijegasint
16

Answer:

sin (90-A) = cos A

tan(90-A) = cot A

Therefore,

= cosA/1-tanA + sinA/1-cotA

= cosA/(cosA-sinA/cosA) + sinA/(sinA-cosA/sinA)

=cos²A/(cosA-sinA) + sin²A/(sinA-cosA)

=(cos²A-sin²A)/(cosA-sinA)

a²-b²=(a+b)(a-b)

={(cosA+sinA)(cosA-sinA)}/(cosA-sinA)

=cosA+sinA

=cosA+cos(90-A)

Hence the proved

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