The expression \tan \rm A \over 1 - \cot \rm A + \cot \rm A \over 1 - \tan \rm A can be written as:

Trigonometric Ratio and Identites PYQ — JEE Main 2013 · Shift 0 · Q45 with Solution

View written solutionFree

Correct answer: B

=\frac{t}{1-1/t}+\frac{1/t}{1-t}.$$ 2. Simplify each term: $$\frac{t}{1-1/t}=\frac{t}{\frac{t-1}{t}}=\frac{t^2}{t-1},$$ and $$\frac{1/t}{1-t}=\frac{1}{t(1-t)}=-\frac{1}{t(t-1)}.$$ Hence, $$\frac{t^2}{t-1}-\frac{1}{t(t-1)}.$$ 3. Take common denominator $t(t-1)$: $$\frac{t^2}{t-1}= rac{t^3}{t(t-1)}.$$ Therefore, $$\frac{t^3}{t(t-1)}- rac{1}{t(t-1)} =\frac{t^3-1}{t(t-1)}.$$ 4. Factorize $t^3-1$: $$t^3-1=(t-1)(t^2+t+1).$$ So, $$\frac{t^3-1}{t(t-1)}=\frac{(t-1)(t^2+t+1)}{t(t-1)}=\frac{t^2+t+1}{t}.$$ Thus, $$\frac{t^2+t+1}{t}=t+1+\frac{1}{t}=\tan A+1+\cot A.$$ 5. Now use $$\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} =\frac{\sin^2 A+\cos^2 A}{\sin A\cos A} =\frac{1}{\sin A\cos A}=\sec A\csc A.$$ Therefore the expression equals $$\tan A+\cot A+1=\sec A\csc A+1.$$ 6. Check options: - A: $\sin A\cos A+1$ ❌ - B: $\sec A\csc A+1$ ✅ - C: $\tan A+\cot A$ ❌ (missing the $+1$) - D: $\sec A+\csc A$ ❌ Hence the correct option is **B**.

View written solutionFree

Correct answer: B

=\frac{t}{1-1/t}+\frac{1/t}{1-t}.$$ 2. Simplify each term: $$\frac{t}{1-1/t}=\frac{t}{\frac{t-1}{t}}=\frac{t^2}{t-1},$$ and $$\frac{1/t}{1-t}=\frac{1}{t(1-t)}=-\frac{1}{t(t-1)}.$$ Hence, $$\frac{t^2}{t-1}-\frac{1}{t(t-1)}.$$ 3. Take common denominator $t(t-1)$: $$\frac{t^2}{t-1}= rac{t^3}{t(t-1)}.$$ Therefore, $$\frac{t^3}{t(t-1)}- rac{1}{t(t-1)} =\frac{t^3-1}{t(t-1)}.$$ 4. Factorize $t^3-1$: $$t^3-1=(t-1)(t^2+t+1).$$ So, $$\frac{t^3-1}{t(t-1)}=\frac{(t-1)(t^2+t+1)}{t(t-1)}=\frac{t^2+t+1}{t}.$$ Thus, $$\frac{t^2+t+1}{t}=t+1+\frac{1}{t}=\tan A+1+\cot A.$$ 5. Now use $$\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A} =\frac{\sin^2 A+\cos^2 A}{\sin A\cos A} =\frac{1}{\sin A\cos A}=\sec A\csc A.$$ Therefore the expression equals $$\tan A+\cot A+1=\sec A\csc A+1.$$ 6. Check options: - A: $\sin A\cos A+1$ ❌ - B: $\sec A\csc A+1$ ✅ - C: $\tan A+\cot A$ ❌ (missing the $+1$) - D: $\sec A+\csc A$ ❌ Hence the correct option is **B**.