Question

prove that sin5x-2sin3x+sinx/cos5x-cosx=cosec2x-cot2x​

Answer 1

Answer:

add sinx and sin5x take common 2sin3x ,cos2x-1 will left in bracket

below after subtraction u will get -2sin3xsin2x

cancel 2sin3xthen solve it

Answer 2

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

LHS

$\tt{\rightarrow\dfrac{(sin5x+sinx)-2sin3x}{cos5x-cosx}}$

$\boxed{\begin{minipage}{11 cm} Formula:- \\ \\ \sinC+sinD=2sin\dfrac{C+D}{2}cos\dfrac{C-D}{2} \\ \\ cosC-cosD =-2sin\dfrac{C+D}{2}sin\dfrac{C-D}{2} \end{minipage}}$

$\tt{\rightarrow\dfrac{2sin3xcos2x-2sin3x}{-2sin3xsin2x}}$

$\tt{\rightarrow\dfrac{2sin3x(cos2x-1)}{-2sin3xsin2x}}$

$\tt{\rightarrow\dfrac{(1-cos2x)}{sin2x}}$

$\tt{\rightarrow\dfrac{1}{sin2x}-\dfrac{cos2x}{sin2x}}$

= cosec2x - cot2x

$\boxed{\begin{minipage}{11 cm} Fundamental Trignometric Indentities \\ \\ \tan (90 - A) = cotA \\ \\ cot (90 - A) = tanA \\ \\ sec (90 - A) = cosecA \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \\ \\ cosec (90 - A) = secA \\ \\ sin^{2}\theta+\cos^{2}\theta =1\\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \end{minipage}}$