Question

if tan3x tan3y= 1 then sin(x+y)=?​

Answer 1

Solution

tan3x tan3y=1

=>tan3x=1/tan3y

=>tan3x=Cot3y

=>tan3x=tan(90°-3y)

=>3x=90°-3y [as, canceling'tan" function]

=>3x+3y=90°

=>3(x+y)=90°

=>(x+y)=90°/3

=>(x+y)=30°

now.......

sin(x+y)=sin30°

=1/2

Answer 2

${\large\bf{\mid{\overline{\underline{Given:-}}}\mid}}$

• Tan3x*tan3y = 1

$\Large\underline\mathfrak{Question}$

• sin(x+y)

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• when TanA×tanB = 1 than,, (A+B) = 90°

$\underline {\underline{\LARGE{{\bf{\green{S}}}{\mathfrak{o}}{\mathfrak{\orange{l}}}{\mathfrak{\red{u}}}{\mathfrak{\pink{t}}}{\mathfrak{\purple{i}}}{\mathfrak{\blue{o}}}{\mathfrak{\red{n}}}}}} : \:$

since ,

Tan3x × tan 3y = 1 = tanA × tanB

comparing we get,

→ A = 3x

→ B = 3y

Hence,

→ 3x + 3y = 90°

→ 3(x+y) = 90°

Dividing both sides by 3 we get,

$\pink{\large\boxed{\boxed{\bold{x + y = 30 \degree}}}}$

_____________________________

Now ,

$\sin(x + y) = \sin(30 \degree) \\ \\ \green{\textbf{we know that sin 30°=}} \: \frac{1}{2}$

$hence \: our \: required \: answer \: is \: \\ \\ \bold{\boxed{\large{\boxed{\orange{\small{\boxed{\large{\red{\bold{\: \frac{1}{2 } }}}}}}}}}}$

$\huge\underline\mathfrak\green{Hope\:it\:Helps\:You}$

#BAL