Question

if sinasinb-cosacosb =1 show that tana+tanb=0​

Answer 1

Step-by-step explanation:

Given:-

Sin A Sin B - Cos A Cos B = 1

To Show :-

Tan A + Tan B = 0

Solution:-

Given that :

Sin A Sin B - Cos A Cos B = 1

It can be written as

-(Cos A Cos B -Sin A Sin B) = 1

We know that

Cos A Cos B - Sin A Sin B = Cos (A+B)

=>- Cos (A+B) = 1

=>Cos (A+B) = -1

=>Cos (A+B) = Cos 180°

=>A+B = 180°

On taking Sin both sides

=>Sin (A+B)= Sin 180°

=>Sin (A+B)= Sin(90°+90°)

We know that Sin (90°+A) = Cos A

=>Sin (A+B)= Cos 90°

=>Sin (A+B)= 0° --------------(1)

Now ,

Tan A + Tan B

=>(Sin A/Cos A ) + (Sin B/Cos B)

=>(Sin A Cos B + Cos A Sin B )/ Cos A Cos B

We know that

Sin A Cos B + Cos A Sin B = Sin (A+B)

=>Sin (A+B)/Cos A Cos B

From (1)

=>0/(Cos A Cos B)

=>0

Tan A + Tan B = 0

Answer:-

If Sin A Sin B - Cos A Cos B = 1 then

Tan A + Tan B = 0

Used formulae:-

• Cos A Cos B - Sin A Sin B = Cos (A+B)

• Sin (90°+A) = Cos A

• Sin A Cos B + Cos A Sin B = Sin (A+B)

Answer 2

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sin a sin b - cos a cos b + 1 = 0

cos a cos b - sin a sin b = 1

cos (a+b) = 1

Hence : sin (a+b) = 0

tan (a+b) = 0/1 = 0

( tan a + tan b ) / ( 1 - tan a tan b ) = 0

tan a + tan b = 0

tan a [ 1 + ( tan b / tan a ) ] = 0

1 + ( 1/ tan a ) tan b = 0

1 + cot a tan b = 0

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