Question

if tan A = cot B than A+B =____​

Answer 1

Answer:

tan²

A is tan and B is cot

we know that tan = cot

Than I am converted cot to tan okay

and there is a+b so tan + tan = tan²

I hope it will help you

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Answer 2

$\textsf{\large{\underline{Solution}:}}$

Given That:

$\rm\longrightarrow \tan A=\cot B$

We know that:

$\rm\longrightarrow tan(90^{\circ}-\alpha)=\cot \alpha$

Therefore:

$\rm\longrightarrow\tan A=\tan(90^{\circ}-B)$

$\rm\longrightarrow A=90^{\circ}-B$

$\rm\longrightarrow A+B=90^{\circ}$

Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

1. Relationship between sides and T-Ratios.

• sin(x) = Height/Hypotenuse
• cos(x) = Base/Hypotenuse
• tan(x) = Height/Base
• cot(x) = Base/Height
• sec(x) = Hypotenuse/Base
• cosec(x) = Hypotenuse/Height

2. Square formulae.

• sin²x + cos²x = 1
• cosec²x - cot²x = 1
• sec²x - tan²x = 1

3. Reciprocal Relationship.

• sin(x) = 1/cosec(x)
• cos(x) = 1/sec(x)
• tan(x) = 1/cot(x)

4. Cofunction identities.

• sin(90° - x) = cos(x)
• cos(90° - x) = sin(x)
• cosec(90° - x) = sec(x)
• sec(90° - x) = cosec(x)
• tan(90° - x) = cot(x)
• cot(90° - x) = tan(x)

5. Even odd identities.

• sin(-x) = -sin(x)
• cos(-x) = cos(x)
• tan(-x) = -tan(x)