If A - B - - 4- then 1-tan A 1-tan B is equal toA- 2B- 1C- 0D- 3

The correct option is A 2 tan(A B)= tan π/4 ⇒tanA tanB/1+tanAtanB =1 ⇒ tanA tanB=1+tanAtanB .....(1) Now, (1+tanA)(1 tanB)=1+tanA tanB tanAtanB =1+1+tanAtanB ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If A - B =π /...

Question

If A-B = $π / 4$ , then (1+tanA)(1-tanB) is equal to

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Solution

The correct option is A 2

tan(A-B)= $t a n \frac{π}{4}$
$\Rightarrow \frac{t a n A - t a n B}{1 + t a n A t a n B} = 1$
$\Rightarrow t a n A - t a n B = 1 + t a n A t a n B$ .....(1)
Now,
(1+tanA)(1-tanB)=1+tanA-tanB-tanAtanB
=1+1+tanAtanB-tanAtanB [Using eq. (1)]
=2

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If A-B = π/4, then (1+tanA)(1-tanB) is equal to

The correct option is A 2 tan(A-B)= tanπ4 ⇒tanA−tanB1+tanAtanB=1 ⇒tanA−tanB=1+tanAtanB .....(1) Now, (1+tanA)(1-tanB)=1+tanA-tanB-tanAtanB =1+1+tanAtanB-tanAtanB [Using eq. (1)] =2

If A-B = $π / 4$ , then (1+tanA)(1-tanB) is equal to

The correct option is A 2

tan(A-B)= $t a n \frac{π}{4}$
$\Rightarrow \frac{t a n A - t a n B}{1 + t a n A t a n B} = 1$
$\Rightarrow t a n A - t a n B = 1 + t a n A t a n B$ .....(1)
Now,
(1+tanA)(1-tanB)=1+tanA-tanB-tanAtanB
=1+1+tanAtanB-tanAtanB [Using eq. (1)]
=2