Hints to prove \(\displaystyle{{\tan}^{{2}}{\left\lbrace\theta\right\rbrace}}={\tan{{\left\lbrace{A}\right\rbrace}}}{\tan{{\left\lbrace{B}\right\rbrace}}}\) ,given \(\displaystyle{\frac{{{\sin{{\left\lbrace{\left(\theta+{A}\right)}\right\rbrace}}}}}{{{\sin{{\left\lbrace{\left(\theta+{B}\right)}\right\rbrace}}}}}}=\sqrt{{{\frac{{{\sin{{\left\lbrace{2}{A}\right\rbrace}}}}}{{{\sin{{\left\lbrace{2}{B}\right\rbrace}}}}}}}}\)

Zack Mora

Zack Mora

Answered question

2022-03-27

Hints to prove tan2{θ}=tan{A}tan{B} ,given sin{(θ+A)}sin{(θ+B)}=sin{2A}sin{2B}

Answer & Explanation

uqhekekocj8f

uqhekekocj8f

Beginner2022-03-28Added 8 answers

sin(θ+A)sin(θ+B)==tanθcosA+sinAtanθcosB+sinB
Dividing numerator & the denominator by cosθ
(tanθcosA+sinA)2(tanθcosB+sinB)2=2sinAcosA2sinBcosB
(tanθ+tanA)2(tanθ+tanB)2=tanAtanB
Dividing numerator of both sides by cos2A
and the denominator of both sides by cos2B
Now simplify assuming tanAtanB

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?