arthham

JEE Advanced 2013 · Trigonometric Functions

The expression tan⁡A1−cot⁡A+cot⁡A1−tan⁡A\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A} can be written as :

Answer: B. sec⁡Acosec⁡A+1\sec A \operatorname{cosec} A + 1

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Written solution

E=tan⁡A1−cot⁡A+cot⁡A1−tan⁡AE = \dfrac{\tan A}{1-\cot A} + \dfrac{\cot A}{1-\tan A} Which option equals E?\text{Which option equals } E? (1) sin⁡Acos⁡A+1(2) sec⁡Acosec⁡A+1\text{(1) } \sin A\cos A+1 \qquad \text{(2) } \sec A\operatorname{cosec}A+1 (3) tan⁡A+cot⁡A(4) sec⁡A+cosec⁡A\text{(3) } \tan A+\cot A \qquad \text{(4) } \sec A+\operatorname{cosec}A cot⁡A=1tan⁡A\cot A = \dfrac{1}{\tan A} tan⁡A1−cot⁡A=tan⁡2Atan⁡A−1\dfrac{\tan A}{1-\cot A} = \dfrac{\tan^{2}A}{\tan A-1} cot⁡A1−tan⁡A=−1tan⁡A (tan⁡A−1)\dfrac{\cot A}{1-\tan A} = \dfrac{-1}{\tan A\,(\tan A-1)} E=tan⁡2Atan⁡A−1−1tan⁡A(tan⁡A−1)E = \dfrac{\tan^{2}A}{\tan A-1} - \dfrac{1}{\tan A(\tan A-1)} E=tan⁡3A−1tan⁡A(tan⁡A−1)E = \dfrac{\tan^{3}A-1}{\tan A(\tan A-1)} tan⁡3A−1=(tan⁡A−1)(tan⁡2A+tan⁡A+1)\tan^{3}A-1 = (\tan A-1)(\tan^{2}A+\tan A+1) E=tan⁡2A+tan⁡A+1tan⁡A=tan⁡A+1+cot⁡AE = \dfrac{\tan^{2}A+\tan A+1}{\tan A} = \tan A + 1 + \cot A tan⁡A+cot⁡A=sin⁡Acos⁡A+cos⁡Asin⁡A=sin⁡2A+cos⁡2Asin⁡Acos⁡A\tan A+\cot A = \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A} = \dfrac{\sin^{2}A+\cos^{2}A}{\sin A\cos A} =1sin⁡Acos⁡A= \dfrac{1}{\sin A\cos A} 1sin⁡Acos⁡A=cosec⁡A sec⁡A\dfrac{1}{\sin A\cos A} = \operatorname{cosec}A\,\sec A E=1+sec⁡Acosec⁡AE = 1 + \sec A\operatorname{cosec}A (2) sec⁡Acosec⁡A+1\boxed{\text{(2) } \sec A\operatorname{cosec}A+1}

More from Trigonometric Functions

All 28 Trigonometric Functions questions →