Question

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

Answer 1

Step-by-step explanation:

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Answer 2

$\\ \\ \underline{ \underline{ \sf \red{solution : } }}$

$\\ \sf \: l.h.s = ( 1 + cot \theta - cosec \theta) \times (1 + tan \theta - sec \theta) \\ \\ \\ \bigstar\boxed{ \bf \:cot \theta = \dfrac{cos \theta}{sin \theta} } \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar\boxed{ \bf \:cosec \theta = \dfrac{1}{sin \theta} } \\ \\ \bigstar \boxed{ \bf \:tan \theta = \dfrac{sin \theta}{cos \theta} } \: \: \: \: \: \: \: \: \: \: \: \: \: \bigstar \boxed{ \bf \: sec \theta = \dfrac{1}{cos \theta} }$

Putting values we get...

$\\ \\ \implies \sf (1 + \dfrac{cos \theta}{sin \theta} - \dfrac{1}{sin \theta} )(1 + \dfrac{sin \theta}{cos \theta} - \dfrac{1}{cos \theta} ) \\ \\ \\ \implies \sf ( \dfrac{sin \theta + cos \theta - 1}{sin \theta} )( \dfrac{cos \theta + sin \theta - 1}{cos \theta} ) \\ \\ \\ \bigstar \boxed{ \bf \:(x + y)(x - y) = {x}^{2} - {y}^{2} } \\ \sf \: here... \to \: x = sin \theta + cos \theta\\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \to \: y = 1 \:$

Putting values we get..

$\\ \\ \sf \implies \dfrac{( {sin \theta + cos \theta)}^{2} - {1}^{2} }{sin \theta.cos \theta} \\ \\ \\ \bigstar\boxed{ \bf \:(x + {y)}^{2} = {x}^{2} + {y}^{2} + 2xy } \\ \sf \: here \: \: \: \: \: \: \to \: x = sin \theta \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \to \: y = cos \theta \\ \\ \sf \: putting \: values \: we \: get... \\ \\ \\ \sf \implies \dfrac{ {sin}^{2} \theta + {cos}^{2} \theta + 2sin \theta.cos \theta - 1 }{sin \theta.cos \theta} \\ \\ \\ \bigstar\boxed{ \bf \: {sin}^{2} \theta + {cos}^{2} \theta = 1 } \\ \\ \\ \sf \implies \dfrac{1 + 2sin \theta.cos \theta - 1}{sin \theta.cos \theta} \\ \\ \\ \sf \implies \dfrac{2 \: \cancel{sin \theta.cos \theta}}{ \cancel{sin \theta.cos \theta} } \\ \\ \\ \sf \implies 2 \: \: = \: \: r.h.s \: \: \: \: \: \: \: \: \: (proved)$

MORE IDENTITIES :-

• 1 + tan²A = sec²A

• 1 + cot²A = cosec²A