Educational

Find the Value of \(8\tan x – \sqrt{5}\sec y\) Question: If \[ \sin x = \frac{3}{5}, \quad \tan y = \frac{1}{2} \] and \[ \frac{\pi}{2} < x < \pi, \quad \pi < y < \frac{3\pi}{2} \] find the value of: \[ 8\tan x – \sqrt{5}\sec y \] Solution Given, \[ \sin x = \frac{3}{5} \] […]

Find the Value of sec x + tan x Question: If \[ \sin x = \frac{12}{13} \] and \(x\) lies in the second quadrant, find the value of: \[ \sec x + \tan x \] Solution Given, \[ \sin x = \frac{12}{13} \] Since \(x\) lies in Quadrant II, sine is positive while cosine and

Find the Values of Other Five Trigonometric Functions Question: Find the values of other five trigonometric functions in the following : \[ \sin x = \frac{3}{5}, \quad x \text{ lies in Quadrant I} \] Solution Given, \[ \sin x = \frac{3}{5} \] Since \(x\) lies in Quadrant I, all trigonometric functions are positive. Using the

Find the Values of Other Five Trigonometric Functions Question: Find the values of other five trigonometric functions in the following : \[ \tan x = \frac{3}{4}, \quad x \text{ lies in Quadrant III} \] Solution Given, \[ \tan x = \frac{3}{4} \] Since \(x\) lies in Quadrant III, both sine and cosine are negative while

Find the Values of Other Five Trigonometric Functions Question: Find the values of other five trigonometric functions in the following : \[ \cos x = -\frac{1}{2}, \quad x \text{ lies in Quadrant II} \] Solution Given, \[ \cos x = -\frac{1}{2} \] Since \(x\) lies in Quadrant II, sine is positive and cosine is negative.

Find the Values of Other Five Trigonometric Functions if cot x = 12/5 in Quadrant III Find the Values of Other Five Trigonometric Functions if cot x = 12/5, x in Quadrant III Question: Find the values of other five trigonometric functions in the following : \[ \cot x = \frac{12}{5}, \quad x \text{ in

If \[ T_n=\sin^n x+\cos^n x \] Prove that \[ 6T_{10}-15T_8+10T_6-1=0 \] Solution: \[ T_6=\sin^6 x+\cos^6 x \] \[ = 1-3\sin^2 x\cos^2 x \] \[ T_8=\sin^8 x+\cos^8 x \] \[ = (\sin^4 x+\cos^4 x)^2-2\sin^4 x\cos^4 x \] \[ = (1-2\sin^2 x\cos^2 x)^2 -2\sin^4 x\cos^4 x \] \[ = 1-4\sin^2 x\cos^2 x+2\sin^4 x\cos^4 x \] \[ T_{10}=\sin^{10}

If \[ T_n=\sin^n x+\cos^n x \] Prove that \[ 2T_6-3T_4+1=0 \] Solution: \[ T_6=\sin^6 x+\cos^6 x \] \[ = (\sin^2 x+\cos^2 x)^3 -3\sin^2 x\cos^2 x(\sin^2 x+\cos^2 x) \] \[ = 1-3\sin^2 x\cos^2 x \] Also, \[ T_4=\sin^4 x+\cos^4 x \] \[ = (\sin^2 x+\cos^2 x)^2 -2\sin^2 x\cos^2 x \] \[ = 1-2\sin^2 x\cos^2 x \]

If \[ T_n=\sin^n x+\cos^n x \] Prove that \[ \frac{T_3-T_5}{T_1} = \frac{T_5-T_7}{T_3} \] Solution: \[ T_3-T_5 = (\sin^3 x+\cos^3 x) -(\sin^5 x+\cos^5 x) \] \[ = \sin^3 x(1-\sin^2 x) + \cos^3 x(1-\cos^2 x) \] \[ = \sin^3 x\cos^2 x + \cos^3 x\sin^2 x \] \[ = \sin^2 x\cos^2 x(\sin x+\cos x) \] \[ = \sin^2

Prove that \[ \sqrt{\frac{1-\sin x}{1+\sin x}} + \sqrt{\frac{1+\sin x}{1-\sin x}} = -\frac{2}{\cos x} \] where \[ \frac{\pi}{2}