Question

A+b=90° then prove √tanAtanB+tanAcotB/sinAsecB_sin^2B/cos^2=tanA

Answer 1

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HERE IS YOUR ANSWER GOES LIKE THIS :


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YOUR QUESTION:

A+B=90° Then prove that √tanAtanB+tanAcotB/sinAsecB_sin^2B/cos^2=tanA


ANSWER:

given,LHS=√tanAtanB+tanAcotB/sinAsecB-sin²B/cos²B

=√tanAtan(90-A)+tanAcot(90-A)/sinAsec(90-A)-sin²(90-A)/cos²A

=√tanAcotA+tanAtanA/sinAcosecA-cos²A/cos²A

=√1+tan²A/1-1

=√tan²A

=tanA


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Answer 2

√tanAtanB+tanAcotB/sinAsecB-sin^2B/cos^2A=tanA ( correct question)

Given:

A+B =90°

To Prove:

√tanAtanB + tanAcotB/sinAsecB - sin^2B/cos^2A=tanA

Proof:

1) In this question we will apply the following property of the trigonometry.

• Sin(90° - ∅) = cos ∅

• cos(90° - ∅) = sin ∅

• cosec(90° - ∅) = sec ∅

• Sec(90° - ∅) = cosec ∅

• tan(90° - ∅) = cot ∅

• cot(90° - ∅) = tan ∅

2) Solution of the question:

• √(tanAtanB + tanAcotB/sinAsecB - sin²B/cos²A)

• √(tanAtan(90° - A) + tanAcot(90° - A)/sinAsec(90° - A) - sin²B/cos²(90° - B))

• √(tanA cotA + tanAtanA/sinAcosecA - sin²B/sin²B)

• √(1 + tan²A / 1 - 1)

• √(sec²A - 1)
• √tan²A
• tanA

LHS = RHS