A-tan B B-tan A- Atan B

The correct option is A True A+tan B B+tan A=1tan A+tan B1tan B+tan A #160; #160; #160; #160; #160; #160; #160; #160; #160; # ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios for Sum of Two Angles

A+tan B B+tan...

Question

$\frac{cot A + tan B}{cot B + tan A} = cot A tan B$

True

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False

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Solution

The correct option is A True
$\frac{cot A + tan B}{cot B + tan A} = \frac{\frac{1}{tan A} + tan B}{\frac{1}{tan B} + tan A}$

$= \frac{\frac{1 + tan A tan B}{tan A}}{\frac{1 + tan A tan B}{tan B}}$

$= \frac{tan B}{tan A}$

$= cot A tan B$

So the relation is $True$

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Similar questions

Q. Evaluate

\frac{cot A + tan B}{cot B + tan A} - cot A tan B .

Q. Prove:

\frac{cot A + tan B}{cot B + tan A} = cot A tan B

Q. Prove the following statements.

\frac{cot A + tan B}{cot B + tan A} = cot A tan B

Q. Prove that :

\frac{c}{a - b} = \frac{tan \frac{A}{2} + tan \frac{B}{2}}{tan \frac{A}{2} - tan \frac{B}{2}}

If A and B are acute angles such that tan $A = \frac{1}{2}, t a n B = \frac{1}{3}$ and tan $(A + B) = \frac{t a n A + t a n B}{1 - t a n A t a n B}$ ,find A + B

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cotA+tanBcotB+tanA=cotAtanB

The correct option is A TruecotA+tanBcotB+tanA=1tanA+tanB1tanB+tanA =1+tanAtanBtanA1+tanAtanBtanB =tanBtanA =cotAtanBSo the relation is True

$\frac{cot A + tan B}{cot B + tan A} = cot A tan B$

The correct option is A True
$\frac{cot A + tan B}{cot B + tan A} = \frac{\frac{1}{tan A} + tan B}{\frac{1}{tan B} + tan A}$

$= \frac{\frac{1 + tan A tan B}{tan A}}{\frac{1 + tan A tan B}{tan B}}$

$= \frac{tan B}{tan A}$

$= cot A tan B$

So the relation is $True$