Question

PrOve tHat (1+ cot ∅ - cos ∅ ) ( 1+ tan∅ + sec∅ ) = 2​

Answer 1

$\large{\underline{\bf{\purple{Correct\:Question:-}}}}$

prove that:-

(1+ cot ∅ - cosec ∅) (1 + tan ∅ + sec∅) = 2.

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$\huge{\underline{\bf{\red{Solution:-}}}}$

LHS :-

$\longmapsto \rm\:\:(1 + cot \theta - cosec \theta)(1 + tan \theta + sec \theta) \\ \\\longmapsto \rm\:(1 + \frac{cos \theta}{sin \theta} - \frac{1}{sin \theta})( 1 + \frac{sin \theta}{cos \theta} + \frac{1}{cos \theta}) \\ \\ \longmapsto \rm\: (\frac{sin \theta + cos \theta - 1}{sin \theta}) \\ \\ \longmapsto \rm\: \frac{[(sin \theta + cos \theta) - 1].[(sin \theta + cos \theta) + 1 ]}{sin \theta \: cos \theta} \\ \\ \longmapsto \rm\:\: \frac{(sin \theta + cos \theta) {}^{2} - 1}{sin \theta \: cos \theta} \\ \\ \longmapsto \rm\:\: \frac{ {sin}^{2} \theta + {cos}^{2} \theta + 2sin \theta \: cos \theta - 1}{sin \theta \: cos \theta} \\ \\\longmapsto \rm\:\: \frac{1 + 2sin \theta \: cos \theta - 1}{sin \theta \: cos \theta} \\ \\ \longmapsto \bf{ \pink{sin {}^{2} \theta + {cos}^{2} \theta = 1 }} \\ \\\longmapsto \rm\:\: \frac{2{ \cancel{sin \theta \: cos \theta}}}{{ \cancel{sin \theta \: cos \theta} }} \\ \\ \longmapsto \bf2 = RHS\:\:$

LHS = RHS

Hence proved

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Answer 2

Correct Question :–

Prove that : (1+ cot ∅ - cosec ∅ ) ( 1+ tan∅ + sec∅ ) = 2

TO PROVE :–

$\\ \: \to \: \: { \bold{ [ 1 + cot(\phi) - cosec( \phi)] [1 + tan( \phi) + sec( \phi) ] = 2 }}$

SOLUTION :–

▪︎ Let's take L.H.S. –

$\\ \: \: \: \: \: \: \: \: { \bold{ =[1 + \cot(\phi) - cosec( \phi)] [1 + \tan( \phi) + \sec( \phi) ] }}$

• We know that –

$\\ \: \: \: \: \implies \: \: \: { \red{ \boxed{ \bold{ \cot(\phi) = \frac{ \cos( \phi) }{ \sin( \phi) } , \tan( \phi) = \frac{ \sin( \phi) }{ \cos( \phi)} \: , \: cosec( \phi) = \frac{1}{ \sin( \phi) } \: \: and \: \: \sec( \phi) = \frac{1}{ \cos( \phi) } }}}}$

• So that –

$\\ \: \: \: \: \: \: \: \: { \bold{ = [1 + \frac{ \cos( \phi) }{ \sin( \phi) } - \frac{1}{ \sin( \phi) } ] [1 + \frac{ \sin( \phi) }{ \cos( \phi) } + \frac{1}{ \cos( \phi) } ] }}$

$\\ \: \: \: \: \: \: \: \: { \bold{ = [ \frac{ \sin( \phi) + \cos( \phi) - 1 }{ \sin( \phi) } ] [ \frac{ \cos( \phi) + \sin( \phi) + 1 }{ \cos( \phi) } ] }}$

• Now using property –

$\\ \: \: \: \implies { \red{ \boxed{\bold{(a + b)(a - b) = {a}^{2} - {b}^{2} }}}}$

$\\ \: \: \: \: \: \: \: \: { \bold{ = [ \frac{[\sin( \phi) + \cos( \phi)]^{2} - 1 }{ \sin( \phi) \cos( \phi) } ] }}$

$\\ \: \: \: \: \: \: \: \: { \bold{ = [ \frac{ \sin^{2} ( \phi) + \cos^{2} ( \phi) + 2 \sin( \phi) \cos( \phi) - 1 }{ \sin( \phi) \cos( \phi) } ] }}$

• Now we are using –

$\\ \: \: \: \implies { \red{ \boxed{\bold{ \ { \sin }^{2}( \phi) + { \cos}^{2}( \phi) = 1 }}}}$

$\\ \: \: \: \: \: \: \: \: { \bold{ = [ \frac{ \cancel1 + 2 \sin( \phi) \cos( \phi) - \cancel1 }{ \sin( \phi) \cos( \phi) } ] }}$

$\\ \: \: \: \: \: \: \: \: { \bold{ = [ \frac{ 2 \cancel {\sin( \phi) \cos( \phi) } }{ \cancel{\sin( \phi) \cos( \phi) } } ] }}$

$\\ \: \: \: \: \: \: \: \: { \bold{ = \: \: \: 2 }}$

$\\ \: \: \: \: \: \: \: \: { \bold{ = \: \: \: R.H.S. \: \: \: \: (Hence \: \: proved)}}$