Advertisements
Advertisements
Question
If x is an acute angle and \[\tan x = \frac{1}{\sqrt{7}}\], then the value of \[\frac{{cosec}^2 x - \sec^2 x}{{cosec}^2 x + \sec^2 x}\] is
Options
3/4
1/2
2
5/4
Advertisements
Solution
3/4
We have:
\[\tan x = \frac{1}{\sqrt{7}}\]
\[ \therefore \tan^2 x = \frac{1}{7}\]
Now, dividing the numerator and the denominator of \[\frac{{cosec}^2 x - \sec^2 x}{{cosec}^2 x + \sec^2 x}\text{ by }{cosec}^2 x:\]
\[\frac{1 - \tan^2 x}{1 + \tan^2 x}\]
\[ = \frac{1 - \frac{1}{7}}{1 + \frac{1}{7}}\]
\[ = \frac{6}{8} = \frac{3}{4}\]
APPEARS IN
RELATED QUESTIONS
If \[\tan x = \frac{b}{a}\] , then find the values of \[\sqrt{\frac{a + b}{a - b}} + \sqrt{\frac{a - b}{a + b}}\].
If \[\sin x + \cos x = m\], then prove that \[\sin^6 x + \cos^6 x = \frac{4 - 3 \left( m^2 - 1 \right)^2}{4}\], where \[m^2 \leq 2\]
Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]
If \[T_n = \sin^n x + \cos^n x\], prove that \[6 T_{10} - 15 T_8 + 10 T_6 - 1 = 0\]
Prove that: tan (−225°) cot (−405°) −tan (−765°) cot (675°) = 0
Prove that
Prove that:
\[\sin\frac{13\pi}{3}\sin\frac{8\pi}{3} + \cos\frac{2\pi}{3}\sin\frac{5\pi}{6} = \frac{1}{2}\]
If \[0 < x < \frac{\pi}{2}\], and if \[\frac{y + 1}{1 - y} = \sqrt{\frac{1 + \sin x}{1 - \sin x}}\], then y is equal to
If x = r sin θ cos ϕ, y = r sin θ sin ϕ and z = r cos θ, then x2 + y2 + z2 is independent of
sin6 A + cos6 A + 3 sin2 A cos2 A =
If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]
The value of sin25° + sin210° + sin215° + ... + sin285° + sin290° is
If sec x + tan x = k, cos x =
The value of \[\tan1^\circ \tan2^\circ \tan3^\circ . . . \tan89^\circ\] is
Find the general solution of the following equation:
Find the general solution of the following equation:
Solve the following equation:
Solve the following equation:
Solve the following equation:
\[\cot x + \tan x = 2\]
Solve the following equation:
\[5 \cos^2 x + 7 \sin^2 x - 6 = 0\]
Solve the following equation:
\[\sin x - 3\sin2x + \sin3x = \cos x - 3\cos2x + \cos3x\]
Solve the following equation:
sin x tan x – 1 = tan x – sin x
Solve the following equation:
\[2^{\sin^2 x} + 2^{\cos^2 x} = 2\sqrt{2}\]
If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.
Write the general solutions of tan2 2x = 1.
Write the number of points of intersection of the curves
In (0, π), the number of solutions of the equation \[\tan x + \tan 2x + \tan 3x = \tan x \tan 2x \tan 3x\] is
If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of x is
Solve the following equations:
`tan theta + tan (theta + pi/3) + tan (theta + (2pi)/3) = sqrt(3)`
Solve the following equations:
2cos 2x – 7 cos x + 3 = 0
Choose the correct alternative:
If sin α + cos α = b, then sin 2α is equal to
Solve 2 tan2x + sec2x = 2 for 0 ≤ x ≤ 2π.
