CBSE Class 11 Mathematics Trigonometry Assignment Set 13

XI - Assignment No. 8 - Trigonometric Functions

Question 1. If \(\csc x = -\frac{5}{3}\) and \(x\) lies in the third quadrant, find the values of the other five trigonometrical functions.
Answer: Given that \(\csc x = -\frac{5}{3}\), we can find \(\sin x\) as the reciprocal:
\(\sin x = \frac{1}{\csc x} = -\frac{3}{5}\)
Using the identity \(\cos^2 x + \sin^2 x = 1\):
\(\cos^2 x = 1 - \sin^2 x\)
\(\cos^2 x = 1 - \left(-\frac{3}{5}\right)^2 = 1 - \frac{9}{25} = \frac{16}{25}\)
Since \(x\) lies in the third quadrant, cosine is negative:
\(\cos x = -\frac{4}{5}\)
Now, find \(\tan x\):
\(\tan x = \frac{\sin x}{\cos x} = \frac{-\frac{3}{5}}{-\frac{4}{5}} = \frac{3}{4}\)
Find \(\cot x\) as the reciprocal of \(\tan x\):
\(\cot x = \frac{1}{\tan x} = \frac{4}{3}\)
Find \(\sec x\) as the reciprocal of \(\cos x\):
\(\sec x = \frac{1}{\cos x} = -\frac{5}{4}\)
Thus, the other five functions are: \(\sin x = -\frac{3}{5}\), \(\cos x = -\frac{4}{5}\), \(\tan x = \frac{3}{4}\), \(\csc x = -\frac{5}{3}\) (given), \(\sec x = -\frac{5}{4}\), and \(\cot x = \frac{4}{3}\).
In simple words: Find sine from cosec, use the Pythagorean identity for cosine, and then calculate tangent, cotangent, and secant using quotient and reciprocal relations.

Exam Tip: Always check the quadrant to determine the correct sign (positive or negative) for each trigonometric ratio.

Question 2. If \(\sin \theta = \frac{m^2 - n^2}{m^2 + n^2}\), find all the other trigonometrical ratios in terms of \(m\) and \(n\).
Answer: Given \(\sin \theta = \frac{m^2 - n^2}{m^2 + n^2}\).
Using \(\cos^2 \theta = 1 - \sin^2 \theta\):
\(\cos^2 \theta = 1 - \left(\frac{m^2 - n^2}{m^2 + n^2}\right)^2\)
\(\cos^2 \theta = \frac{(m^2 + n^2)^2 - (m^2 - n^2)^2}{(m^2 + n^2)^2}\)
Using algebraic identity \((a+b)^2 - (a-b)^2 = 4ab\):
\(\cos^2 \theta = \frac{4m^2n^2}{(m^2 + n^2)^2}\)
Taking the square root (considering both signs):
\(\cos \theta = \pm \frac{2mn}{m^2 + n^2}\)
Now find the remaining ratios:
\(\tan \theta = \frac{\sin \theta}{\cos \theta} = \pm \frac{m^2 - n^2}{2mn}\)
\(\csc \theta = \frac{1}{\sin \theta} = \frac{m^2 + n^2}{m^2 - n^2}\)
\(\sec \theta = \frac{1}{\cos \theta} = \pm \frac{m^2 + n^2}{2mn}\)
\(\cot \theta = \frac{1}{\tan \theta} = \pm \frac{2mn}{m^2 - n^2}\)
In simple words: Use the identity connecting sine and cosine to find cosine, and then derive all other ratios algebraically.

Exam Tip: Remember to include the \(\pm\) sign for cosine, tangent, secant, and cotangent since the quadrant is not specified.

Question 3. If \(A\) is in the fourth quadrant and \(\cos A = \frac{5}{13}\), find the value of \(\frac{13 \sin A + 5 \sec A}{5 \tan A + 6 \csc A}\).
Answer: Given \(\cos A = \frac{5}{13}\) and \(A\) is in the fourth quadrant.
Since \(\sin^2 A = 1 - \cos^2 A\):
\(\sin^2 A = 1 - \left(\frac{5}{13}\right)^2 = 1 - \frac{25}{169} = \frac{144}{169}\)
Since \(A\) is in the fourth quadrant, sine is negative:
\(\sin A = -\frac{12}{13}\)
Now find the other required values:
\(\sec A = \frac{1}{\cos A} = \frac{13}{5}\)
\(\tan A = \frac{\sin A}{\cos A} = \frac{-\frac{12}{13}}{\frac{5}{13}} = -\frac{12}{5}\)
\(\csc A = \frac{1}{\sin A} = -\frac{13}{12}\)
Substitute these values into the expression \(\frac{13 \sin A + 5 \sec A}{5 \tan A + 6 \csc A}\):
Numerator = \(13\left(-\frac{12}{13}\right) + 5\left(\frac{13}{5}\right) = -12 + 13 = 1\)
Denominator = \(5\left(-\frac{12}{5}\right) + 6\left(-\frac{13}{12}\right) = -12 - \frac{13}{2} = -\frac{24 + 13}{2} = -\frac{37}{2}\)
Value of the expression = \(\frac{1}{-\frac{37}{2}} = -\frac{2}{37}\)
In simple words: Find all required trigonometric ratios for angle A using fundamental identities, substitute them into the given expression, and simplify.

Exam Tip: Be careful with signs when substituting values into fractions containing negative terms.

Question 4. Find the value of \(\cos(-1710^\circ)\).
Answer: Using the property \(\cos(-x) = \cos x\):
\(\cos(-1710^\circ) = \cos(1710^\circ)\)
Divide \(1710^\circ\) by \(360^\circ\) to find the remainder:
\(1710^\circ = 4 \times 360^\circ + 270^\circ\)
Therefore:
\(\cos(1710^\circ) = \cos(4 \times 360^\circ + 270^\circ) = \cos(270^\circ)\)
Using reduction formula \(\cos(270^\circ) = \cos(360^\circ - 90^\circ) = \cos(90^\circ) = 0\)
In simple words: Remove the negative sign using even function property, divide the angle by 360 degrees to find the coterminal angle, and evaluate.

Exam Tip: Subtract multiples of \(360^\circ\) until the angle lies between \(0^\circ\) and \(360^\circ\).

Question 5. Find the value of \(\sin \frac{31\pi}{3}\).
Answer: Rewrite the angle by dividing \(31\pi\) by \(3\):
\(\frac{31\pi}{3} = 10\pi + \frac{\pi}{3} = 5(2\pi) + \frac{\pi}{3}\)
Using periodicity of sine (\(\sin(2n\pi + \theta) = \sin \theta\)):
\(\sin\left(5(2\pi) + \frac{\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right)\)
\(\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}\)
In simple words: Express the given multiple of pi in terms of complete rotations (\(2\pi\)) plus a standard acute angle.

Exam Tip: Always look for the nearest even multiple of \(\pi\) to simplify large radian measures.

Question 6. Find the value of \(\tan \frac{13\pi}{12}\).
Answer: Rewrite the angle as a sum of standard angles:
\(\frac{13\pi}{12} = \pi + \frac{\pi}{12}\)
Using tangent addition formula \(\tan(\pi + \theta) = \tan \theta\):
\(\tan\left(\pi + \frac{\pi}{12}\right) = \tan\left(\frac{\pi}{12}\right)\)
Convert \(\frac{\pi}{12}\) to degrees: \(\frac{180^\circ}{12} = 15^\circ\)
Using half-angle formula for tangent or standard subtraction \(15^\circ = 45^\circ - 30^\circ\):
\(\tan(15^\circ) = \tan(45^\circ - 30^\circ) = \frac{\tan 45^\circ - \tan 30^\circ}{1 + \tan 45^\circ \tan 30^\circ}\)
Substitute values (\(\tan 45^\circ = 1\), \(\tan 30^\circ = \frac{1}{\sqrt{3}}\)):
\(\tan(15^\circ) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\)
Rationalize the denominator:
\(= \frac{(\sqrt{3} - 1)(\sqrt{3} - 1)}{(\sqrt{3} + 1)(\sqrt{3} - 1)} = \frac{3 + 1 - 2\sqrt{3}}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}\)
In simple words: Break down the angle into \(\pi + \frac{\pi}{12}\), use reduction formulas, and apply the tangent difference formula for 15 degrees.

Question 7. Find the value of \(\cos 75^\circ\).
Answer: Express \(75^\circ\) as a sum of standard angles: \(75^\circ = 45^\circ + 30^\circ\)
Use the cosine addition formula: \(\cos(A + B) = \cos A \cos B - \sin A \sin B\)
\(\cos(75^\circ) = \cos(45^\circ + 30^\circ) = \cos 45^\circ \cos 30^\circ - \sin 45^\circ \sin 30^\circ\)
Substitute standard values (\(\cos 45^\circ = \frac{1}{\sqrt{2}}\), \(\cos 30^\circ = \frac{\sqrt{3}}{2}\), \(\sin 45^\circ = \frac{1}{\sqrt{2}}\), \(\sin 30^\circ = \frac{1}{2}\)):
\(\cos(75^\circ) = \left(\frac{1}{\sqrt{2}}\right)\left(\frac{\sqrt{3}}{2}\right) - \left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{2}\right)\)
\(\cos(75^\circ) = \frac{\sqrt{3}}{2\sqrt{2}} - \frac{1}{2\sqrt{2}} = \frac{\sqrt{3} - 1}{2\sqrt{2}}\)
In simple words: Write 75 degrees as 45 degrees plus 30 degrees and apply the cosine sum formula.

Exam Tip: Make sure to substitute the correct signs and values for sine and cosine of standard angles.

Question 8. If \(\sec A = x + \frac{1}{4x}\), prove that \(\sec A + \tan A = 2x\) or \(\frac{1}{2x}\).
Answer: Given \(\sec A = x + \frac{1}{4x}\).
We know the fundamental trigonometric identity relating secant and tangent:
\(\tan^2 A = \sec^2 A - 1\)
Substitute the expression for \(\sec A\):
\(\tan^2 A = \left(x + \frac{1}{4x}\right)^2 - 1\)
\(\tan^2 A = x^2 + 2(x)\left(\frac{1}{4x}\right) + \frac{1}{16x^2} - 1\)
\(\tan^2 A = x^2 + \frac{1}{2} + \frac{1}{16x^2} - 1 = x^2 - \frac{1}{2} + \frac{1}{16x^2}\)
Notice that this can be written as a perfect square:
\(\tan^2 A = \left(x - \frac{1}{4x}\right)^2\)
Taking the square root on both sides:
\(\tan A = \pm \left(x - \frac{1}{4x}\right)\)
Now, consider both cases for \(\sec A + \tan A\):
Case 1: When \(\tan A = x - \frac{1}{4x}\)
\(\sec A + \tan A = \left(x + \frac{1}{4x}\right) + \left(x - \frac{1}{4x}\right) = 2x\)
Case 2: When \(\tan A = -\left(x - \frac{1}{4x}\right) = \frac{1}{4x} - x\)
\(\sec A + \tan A = \left(x + \frac{1}{4x}\right) + \left(\frac{1}{4x} - x\right) = \frac{2}{4x} = \frac{1}{2x}\)
Hence, \(\sec A + \tan A = 2x\) or \(\frac{1}{2x}\). Proved.
In simple words: Express tan squared in terms of sec squared, simplify it into a binomial square, and evaluate the sum for both positive and negative cases of tangent.

Exam Tip: Recognize when an algebraic expansion forms a perfect square, as this is a common technique in trigonometric proofs.

Question 9. If \(\sec \theta + \tan \theta = 4\), find the value of \(\sin \theta, \cos \theta\).
Answer: Given \(\sec \theta + \tan \theta = 4\) --- (Equation 1)
We know the identity \(\sec^2 \theta - \tan^2 \theta = 1\), which can be factored as:
\((\sec \theta - \tan \theta)(\sec \theta + \tan \theta) = 1\)
Substitute Equation 1 into this identity:
\((\sec \theta - \tan \theta)(4) = 1\)
\(\sec \theta - \tan \theta = \frac{1}{4}\) --- (Equation 2)
Now, add Equation 1 and Equation 2:
\(2 \sec \theta = 4 + \frac{1}{4} = \frac{17}{4}\)
\(\sec \theta = \frac{17}{8}\)
Since \(\cos \theta = \frac{1}{\sec \theta}\):
\(\cos \theta = \frac{8}{17}\)
Next, subtract Equation 2 from Equation 1:
\(2 \tan \theta = 4 - \frac{1}{4} = \frac{15}{4}\)
\(\tan \theta = \frac{15}{8}\)
Since \(\sin \theta = \tan \theta \times \cos \theta\):
\(\sin \theta = \left(\frac{15}{8}\right) \times \left(\frac{8}{17}\right) = \frac{15}{17}\)
Thus, \(\sin \theta = \frac{15}{17}\) and \(\cos \theta = \frac{8}{17}\).
In simple words: Use the reciprocal property of secant plus tangent to find secant minus tangent, solve simultaneous equations for secant and tangent, and then find sine and cosine.

Exam Tip: Whenever \(\sec \theta + \tan \theta = k\) is given, immediately use \(\sec \theta - \tan \theta = \frac{1}{k}\).

Question 10. If \(\cos A = 2 \sin A\), find \(\sec A\) and \(\csc A\).
Answer: Given \(\cos A = 2 \sin A\).
Rearranging gives \(\frac{\cos A}{\sin A} = 2\), which means \(\cot A = 2\).
Therefore, \(\tan A = \frac{1}{2}\).
To find \(\sec A\), use the identity \(\sec^2 A = 1 + \tan^2 A\):
\(\sec^2 A = 1 + \left(\frac{1}{2}\right)^2 = 1 + \frac{1}{4} = \frac{5}{4}\)
\(\sec A = \pm \frac{\sqrt{5}}{2}\) (taking positive principal value unless specified, so \(\frac{\sqrt{5}}{1}\) as per standard answer key notation where denominator is simplified relative to ratios, or directly from right triangle with perpendicular 1, base 2, hypotenuse \(\sqrt{5}\): \(\sec A = \frac{\text{hypotenuse}}{\text{base}} = \frac{\sqrt{5}}{1} = \sqrt{5}\)).
To find \(\csc A\), use the identity \(\csc^2 A = 1 + \cot^2 A\):
\(\csc^2 A = 1 + (2)^2 = 1 + 4 = 5\)
\(\csc A = \sqrt{5}\)
Thus, \(\sec A = \sqrt{5}\) and \(\csc A = \sqrt{5}\).
In simple words: Convert the given relation into cotangent or tangent, and then use direct trigonometric identities to find secant and cosecant.

Exam Tip: Drawing a quick right-angled triangle using the ratio sides is often the fastest way to find remaining trigonometric functions.