Question

Prove that (1 + cot – cosec) (1 + tan + sec) = 2.​

Answer 1

$\sf\begin{gathered}(1+ cot A - cosec A)(1+tanA+secA)=2 \\ L.H.S. \\ =(1+cotA-cosecA)(1+tanA+secA) \\ = (1 + \frac{cosA}{sinA} - \frac{1}{sinA} )(1+ \frac{sinA}{cosA} + \frac{1}{cosA} ) \\ = ( \frac{sinA+cosA-1}{sinA} )( \frac{cosA+sinA+1}{cosA} ) \\ = \frac{ (sinA+cosA)^{2}- 1^{2} }{sinA.cosA} \\ = \frac{ sin^{2}A + cos^{2}A + 2sinA.cosA-1 }{sinA.cosA} \\ = \frac{ 1 + 2sinA.cosA-1 }{sinA.cosA} \\ = \frac{2sinA.cosA }{sinA.cosA} = 2 = R.H.S.\end{gathered}$

Hence proved

Answer 2

$\begin{gathered}\sf\begin{gathered}(1+ cot A - cosec A)(1+tanA+secA)=2 \\ L.H.S. \\ =(1+cotA-cosecA)(1+tanA+secA) \\ = (1 + \frac{cosA}{sinA} - \frac{1}{sinA} )(1+ \frac{sinA}{cosA} + \frac{1}{cosA} ) \\ = ( \frac{sinA+cosA-1}{sinA} )( \frac{cosA+sinA+1}{cosA} ) \\ = \frac{ (sinA+cosA)^{2}- 1^{2} }{sinA.cosA} \\ = \frac{ sin^{2}A + cos^{2}A + 2sinA.cosA-1 }{sinA.cosA} \\ = \frac{ 1 + 2sinA.cosA-1 }{sinA.cosA} \\ = \frac{2sinA.cosA }{sinA.cosA} = 2 = R.H.S.\end{gathered} \end{gathered}$