Question

Match the following
$\begin{array}{cccc}& ListI& & ListII\\ A.& \frac{1+cosA-sinA}{1+cosA+sinA}& i.& \frac{1}{cosecA}+\frac{1}{secA}\\ B.& (1+\frac{1}{ta{n}^{2}A})(1+\frac{1}{co{t}^{2}A})& II.& secA-tanA\\ C.& \frac{cosA}{1-tanA}+\frac{sinA}{1-cotA}& iii.& \frac{1}{co{s}^{2}A-cos4A}\end{array}$

• A. A-ii, B-iii, C-i
• B. A-iii, B- ii, C-i
• C. A- i, B-ii, C-iii
• D. None of these

Answer 1

The correct option is A

A-ii, B-iii, C-i

A)
$\frac{1+cosA-sinA}{1+cosA+sinA}=\frac{secA+1-tanA}{secA+1+tanA}$
[dividing numerator and denominator by $cos\theta$]
$=\frac{[secA-tanA+se{c}^{2}A-ta{n}^{2}A]}{secA+tanA+1}=\frac{[secA-tanA+(secA-tanA\left)\right(secA+tanA\left)\right]}{secA+tanA+1}=\frac{\left[\right(secA-tanA\left)\right(1+secA+tanA\left)\right]}{secA+tanA+1}=secA-tanA$

B)
$\left(\frac{ta{n}^{2}A+1}{ta{n}^{2}A}\right)\left(\frac{co{t}^{2}A+1}{co{t}^{2}A}\right)=\left(\frac{se{c}^{2}A}{ta{n}^{2}A}\right)\left(\frac{cose{c}^{2}A}{co{t}^2}\right)=\left(\frac{co{s}^{2}A}{co{s}^{2}A\times si{n}^{2}A}\right)\left(\frac{si{n}^{2}A}{si{n}^{2}A\times co{s}^{2}A}\right)=\left(\frac{1}{si{n}^{2}A}\right)\left(\frac{1}{co{s}^{2}A}\right)$

C. Similarly. do yourself
$\frac{cosA}{1-tanA}+\frac{sinA}{1-cotA}=\frac{1}{cosecA}+\frac{1}{secA}$