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