Question

Sin5A - sin3A / cos3A + cos5A = tanA

Answer 1

Answer:

$\frac{sin5A-sin3A}{cos3A+cos5A}=tanA$

Step-by-step explanation:

$LHS=\frac{sin5A-sin3A}{cos3A+cos5A}$

$=\frac{2cos\left(\frac{5A+3A}{2}\right) sin\left(\frac{5A-3A}{2}\right)}{2cos\left(\frac{3A+5A}{2}\right) cos\left(\frac{3A-5A}{2}\right)}$

$=\frac{cos\left(\frac{8A}{2}\right) sin\left(\frac{2A}{2}\right)}{cos\left(\frac{8A}{2}\right) cos\left(\frac{-2A}{2}\right)}$

$=\frac{cos\left(\frac{8A}{2}\right) sin\left(\frac{2A}{2}\right)}{cos\left(\frac{8A}{2}\right) cos\left(\frac{2A}{2}\right)}\\=\frac{cos4A sinA}{cos4A cosA}\\=\frac{sinA}{cosA}=tanA=RHS$

Therefore,

$\frac{sin5A-sin3A}{cos3A+cos5A}=tanA$

•••♪

Answer 2

$\huge{\red{\underline{\pink{\mathbb{HeY, mAtE..!!!}}}}}$

See the attachment✌☺_______✔