Question

IF tan AtanB = tan 45° so; Sin (A+B) = ?​

Answer 1

We have

CosACosB- SinASinB =0

Cos (A+B) =0

Now, Sin (A +B) =square root of [1 - Cos ^2 (A+B)]

= 1

Answer 2

the value of sin(A +B ) = ½

Step-by-step explanation:

$\\ \sf \large \red{Given}$

• tanA + tanB = tan45°

$\\ \sf \large \red{To \: find}$

• the value of Sin (A+B) .

$\\ \sf \large \red {Solution}$

$\sf \: tanA + tanB = tan 45 \degree \\ \\ \sf \: tan(A +B ) =tan 45 \degree$

tan is taken as common .

$\sf \: \cancel{tan} \: (A +B ) = \cancel{tan } \: 45 \degree \\ \\ \boxed{ \sf \: A +B = 45 \degree}$

Now, value of sin ( A + B ) :-

we have find out the value of A + B = 45° .

So,

$\underline{\boxed{\sf \:sin 45\degree \: = \dfrac{1}{ \sqrt{2} }}} \\ \\ \sf \: Here, \: we \: have \: put \: the \: value \: of \:A \: and \: B\:$

Hence, the value of sin(A +B ) = ½

Trigonometric tabe for some angles :-

$\Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 65^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$