\(cos^{-1}(\frac{y}{2})=log_e(\frac{x}{5})^5, |y| < 2\)
Differentiate on both side
\(-\frac{ 1}{\sqrt{1-(\frac{y}{2})^2}} \times \frac{y'}{2} =\frac{ 5}{\frac{x}{5}} \times \frac{1}{5}\)
\(-\frac{xy'}{2} = 5 \sqrt{1-(\frac{y}{2})^2}\)
Square on both side
\(\frac{x^2y'^2}{4} = 25\bigg(\frac{4-y^2}{4}\bigg)\)
Diff on both side
\(2xy'^2+2y'y''x^2=-25 \times 2yy'\)
\(xy'+y''x^2+25y = 0\)
Hence, the correct option is (D): \(x^2y′′ + xy′+ 25y = 0\)
Given that $\sin \theta + \cos \theta = x$, prove that $\sin^4 \theta + \cos^4 \theta = \dfrac{2 - (x^2 - 1)^2}{2}$.
The term independent of $ x $ in the expansion of $$ \left( \frac{x + 1}{x^{3/2} + 1 - \sqrt{x}} \cdot \frac{x + 1}{x - \sqrt{x}} \right)^{10} $$ for $ x>1 $ is:
Various trigonometric identities are as follows:
Cosecant and Secant are even functions, all the others are odd.
T-Ratios of (2x)
sin2x = 2sin x cos x
cos 2x = cos2x – sin2x
= 2cos2x – 1
= 1 – 2sin2x
T-Ratios of (3x)
sin 3x = 3sinx – 4sin3x
cos 3x = 4cos3x – 3cosx