Question

Prove that :-
cot4x(sin5x + sin3x) = cotx ( sin5x - sin3x)

Answer 1

LHS=cot4x (sin5x+sin3x)
=cot4x.2sin4x.cosx
=2(cos4x/sin4x)sin4x. cosx
=2co4x.cosx
=2cos4x. sinx.cotx
=cotx(2sinx.cos4x)
=cotx(sin5x-sin3x)=RHS

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Taking Left hand side

cot4x(sin5x + sin3x)

$\tt{\rightarrow sinA+sinB=2sin\dfrac{A+B}{2}cos\dfrac{A-B}{2}}$

$\tt{\rightarrow\dfrac{cos4x}{sin4x}{2sin(\dfrac{5x+3x}{2})cos(\dfrac{5x-3x}{2})}}$

$\tt{\rightarrow\dfrac{cos4x}{sin4x}{2sin4xcosx}}$

= 2cos4xcosx

Now,

Right Hand side

cotx(sin5x - sin3x)

$\tt{\rightarrow sinA-sinB=2cos\dfrac{A+B}{2}sin\dfrac{A-B}{2}}$

$\tt{\rightarrow\dfrac{cosx}{sinx}{2cos4xsinx}}$

= 2cos4xcosx

Hence Proved

LHS = RHS

$\boxed{\begin{minipage}{9 cm} Fundamental Trignometric Indentities \\ \\ \sin^{2}\theta+\cos^{2}\theta =1 \\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \end{minipage}}$