If cos θ = \(\frac{5}{13}\) , what is the value of cot θ ?
The correct answer is \frac{5}{12}.
Given \cos \theta = \frac{5}{13} , we could find \sin \theta using \sin^2 \theta + \cos^2 \theta = 1 : \sin^2 \theta + \left(\frac{5}{13}\right)^2 = 1 \sin^2 \theta + \frac{25}{169} = 1 \sin^2 \theta = 1 - \frac{25}{169} = \frac{169 - 25}{169} = \frac{144}{169} \sin \theta = \sqrt{\frac{144}{169}} = \frac{12}{13} (assuming θ is in a quadrant where sin is positive) Then, use the identity \cot \theta = \frac{\cos \theta}{\sin \theta} : \cot \theta = \frac{5/13}{12/13} = \frac{5}{13} \times \frac{13}{12} = \frac{5}{12} This method also gives the same result for \cot \theta , confirming our previous calculation using the triangle side ratios.
Similar Questions
If tan 3 θ = cot(θ - 22°), where 3 θ is an angle, find the value of θ.
The value of cot (-315°) is :
The value of \(\frac{\tan^2 (22^\circ - \theta)-\tan (\theta + 68^\circ)\ -\ \text{cosec}^2 (68^\circ + \theta)+\cot (22^\circ - \theta)} {3 (\cot^2 52^\circ - \sec^2 38^\circ)+ 2 (\text{cosec}^2 \ 28^\circ - \tan^2 62^\circ)}\) is:
The value of \(\frac{{{{\sec }^2}60^\circ {{\cos }^2}45^\circ \, + \,{\rm{cose}}{{\rm{c}}^2}30^\circ }}{{{{\sec }^2}45^\circ - \tan 45^\circ }}\) is:
The value of \(\sin^2\frac{2\pi}{3}+\cos^2\frac{5\pi}{6}-\tan^2\frac{3\pi}{4}\) is :
sinθ + cosec θ = 2, then the value of sin 99 θ + cosec 99 θ is:
Determine the largest angle if the angles of a triangle are in the ratio 3 : 4 : 5.
The value of the expression \(\rm \frac{4\sin^230^{\circ}+\cos^260^{\circ}-\tan^245^{\circ}}{2\sin60^{\circ}\cos30^{\circ}-\tan45^{\circ}}\) is:
The general solution of the equation \(\tan 3x + \cot (2x + \frac{\pi }{3}) = 0\) is:
If tan 15θ = cot 15θ (0° < θ < 10°) , then the value of θ is:
Important Questions from Circular Measure of Angles
Which of the following values of A and B satisfies, sin(A + B) = sin A + sin B, where
If tan 3 θ = cot(θ - 22°), where 3 θ is an angle, find the value of θ.
The value of 4cos\(\left( {\frac{\pi }{6}\, - \,\alpha } \right)\) sin\(\left( {\frac{\pi }{3}\, - \,\alpha } \right)\) is equal to:
1 + tan 15° cot 75° is equal to:
If tan θ = 1/√5, find the value of cosec2θ – sec2θ.
