Advertisements
Advertisements
Question
Prove the:
\[ \sqrt{\frac{1 - \sin x}{1 + \sin x}} + \sqrt{\frac{1 + \sin x}{1 - \sin x}} = - \frac{2}{\cos x},\text{ where }\frac{\pi}{2} < x < \pi\]
Advertisements
Solution
\[LHS = \left| \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right| + \left| \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right|\]
\[ = \left| \sqrt{\frac{\left( 1 - \sin x \right)\left( 1 - \sin x \right)}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)}} \right| + \left| \sqrt{\frac{\left( 1 + \sin x \right)\left( 1 + \sin x \right)}{\left( 1 - \sin x \right)\left( 1 + \sin x \right)}} \right|\]
\[ = \left| \sqrt{\frac{\left( 1 - \sin x \right)\left( 1 - \sin x \right)}{\left( 1 + \sin x \right)\left( 1 - \sin x \right)}} \right| + \left| \sqrt{\frac{\left( 1 + \sin x \right)\left( 1 + \sin x \right)}{\left( 1 - \sin x \right)\left( 1 + \sin x \right)}} \right|\]
\[ = \left| \sqrt{\frac{\left( 1 - \sin x \right)^2}{1 - \sin^2 x}} \right| + \left| \sqrt{\frac{\left( 1 + \sin x \right)^2}{1 - \sin^2 x}} \right|\]
\[ = \left| \sqrt{\frac{\left( 1 - \sin x \right)^2}{\cos^2 x}} \right| + \left| \sqrt{\frac{\left( 1 + \sin x \right)^2}{\cos^2 x}} \right|\]
\[ = \left| \frac{1 - \sin x}{\cos x} \right| + \left| \frac{1 + \sin x}{\cos x} \right|\]
\[ = \left| \frac{1 - \sin x + 1 + \sin x}{\cos x} \right|\]
\[ = \left| \frac{2}{\cos x} \right|\]
\[ = - \frac{2}{\cos x} \left[ \because \frac{\pi}{2} < x < \pi \text{ and in the second quadrant, }\cos x \text{ is negative }\right]\]
= RHS
Hence proved .
APPEARS IN
RELATED QUESTIONS
If \[\tan x = \frac{a}{b},\] show that
If \[cosec x - \sin x = a^3 , \sec x - \cos x = b^3\], then prove that \[a^2 b^2 \left( a^2 + b^2 \right) = 1\]
If \[a = \sec x - \tan x \text{ and }b = cosec x + \cot x\], then shown that \[ab + a - b + 1 = 0\]
Prove that: tan 225° cot 405° + tan 765° cot 675° = 0
Prove that
Find x from the following equations:
\[cosec\left( \frac{\pi}{2} + \theta \right) + x \cos \theta \cot\left( \frac{\pi}{2} + \theta \right) = \sin\left( \frac{\pi}{2} + \theta \right)\]
Prove that:
\[\tan 4\pi - \cos\frac{3\pi}{2} - \sin\frac{5\pi}{6}\cos\frac{2\pi}{3} = \frac{1}{4}\]
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 tan \[x = - \frac{1}{\sqrt{5}}\] and θ lies in the IV quadrant, then the value of cos x is
If \[cosec x - \cot x = \frac{1}{2}, 0 < x < \frac{\pi}{2},\]
sin2 π/18 + sin2 π/9 + sin2 7π/18 + sin2 4π/9 =
If tan θ + sec θ =ex, then cos θ equals
Which of the following is correct?
Find the general solution of the following equation:
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:
Solve the following equation:
\[5 \cos^2 x + 7 \sin^2 x - 6 = 0\]
If secx cos5x + 1 = 0, where \[0 < x \leq \frac{\pi}{2}\], find the value of x.
Write the number of solutions of the equation
\[4 \sin x - 3 \cos x = 7\]
If cos x = k has exactly one solution in [0, 2π], then write the values(s) of k.
Write the values of x in [0, π] for which \[\sin 2x, \frac{1}{2}\]
and cos 2x are in A.P.
If \[2 \sin^2 x = 3\cos x\]. where \[0 \leq x \leq 2\pi\], then find the value of x.
The smallest value of x satisfying the equation
A solution of the equation \[\cos^2 x + \sin x + 1 = 0\], lies in the interval
A value of x satisfying \[\cos x + \sqrt{3} \sin x = 2\] is
If \[\sqrt{3} \cos x + \sin x = \sqrt{2}\] , then general value of x is
Find the principal solution and general solution of the following:
tan θ = `- 1/sqrt(3)`
Solve the following equations:
sin 5x − sin x = cos 3
Choose the correct alternative:
`(cos 6x + 6 cos 4x + 15cos x + 10)/(cos 5x + 5cs 3x + 10 cos x)` is equal to
If sin θ and cos θ are the roots of the equation ax2 – bx + c = 0, then a, b and c satisfy the relation ______.
Find the general solution of the equation sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x
In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is ______.
