Question

(1+cotA-cosecA) (1+tanA+secA) = 2​

Answer 1

$\large \underline{ \purple{ \sf{question : }}}$

$\sf{prove \: that : } \\ \sf{(1 + cotA - cosecA )(1 + tanA + secA) = 1}$

$\large \underline{ \blue{ \sf{formulae : }}}$

$\sf{tan \: \theta = \frac{sin \theta}{cos \theta} } \\ \\ \sf{cot \: \theta = \frac{cos \theta}{sin \theta} } \\ \\ \sf{cosec \theta = \frac{1}{sin \theta} } \\ \\ \sf{sec \theta = \frac{1}{cos \theta} }\\ \\ \sf{sin {}^{2} \theta + cos {}^{2} \theta = 1 } \\ \\ \sf{(a - b)(a + b) = a {}^{2} - b {}^{2} }$

$\large \underline{\pink{ \sf{solution :}}}$

$\sf{(1 + cotA - cosecA )(1 + tanA + secA) = 1}$

$\sf{taking \: LHS}$

$\sf{= (1 +cot A - cosecA)(1 + tanA + secA} \\ \\ \sf{= (1 + \frac{cosA}{sinA} - \frac{1}{sinA} })(1 + \frac{sinA}{cosA } + \frac{1}{cosA} ) \\ \\ \sf{ = ( \frac{sinA + cosA - 1}{sinA})( \frac{sinA + cosA + 1}{cosA} )} \\ \\ \sf{ = \frac{(sin A + cos A) {}^{2} - 1 {}^{2} }{sinA.cosA} } \\ \\ \sf{ = \frac{sin {}^{2}A + cos {}^{2}A + 2sin A.cos A - 1}{sinA.cosA} } \\ \\ \sf{ = \frac{1 + 2sinA.cosA - 1}{sinA.cosA} } \\ \\ \sf{= \frac{ \cancel1 + 2sinA.cosA - \cancel 1}{sinA.cosA} } \\ \\ \sf{ = \frac{2sinA.cosA}{sinA.cosA} } \\ \\ \sf{ = \frac{2 \cancel{sinA.cosA}}{ \cancel{sinA.cosA} }} \\ \\ \sf{ = 2} \\ \\ \sf{ʟʜs = ʀʜs} \\ \\ \underline{\red{\sf{hence \: proved}}}$