Question

if
$sin(a + b) = 1$
and
$tan (a - b) = 1 \div \sqrt{3}$
then find the value of Tan A + Cot B

Answer 1

Hii friend,

Sin(A+B)= 1

Sin(A+B) = Sin90°

A+B = 90.....(1)

Tan(A-B) = 1/✓3

Tan(A-B) = Tan30°

(A-B) = 30.......(2)

From equation 1 we get,

A+B= 90


A = 90-B......(3)


Putting the value of A in equation (2)

A - B = 30

90-B -B = 30

-2B = 30-90

-2B = -60

B = -60/-2 = 30

Putting the value of B in equation (3)

A = 90-B => 90-30 = 60°

TanA = Tan 60° = ✓3


and,

CotB = Cot30° = ✓3


Therefore,

TanA + CotA = ✓3 + ✓3 = 2✓3


HOPE IT WILL HELP YOU....... :-)

Answer 2

Hey!

Sin(a + b ) = 1

Sin ( a + b ) = Sin90°

[ Sin90° = 1 ]

a + b = 90 ...... ( I )

tan ( a - b ) = 1 ÷ √3

tan ( a - b ) = tan30°

a - b = 30 ..... ( ii )


Adding ( i ) and ( ii )

a + b + a - b = 90 + 30

2a = 120

a = 60

Putting value of a in ( i )

a + b = 90

60 + b = 90

b = 90 - 60

b = 30

Now ,

We have to find value of
tan a + cotb

tan 60° + cot 30°

$\sqrt{3} + \sqrt{3}$


= 2√3