Question

19. Prove that:(1 + cot0 - cosec 0) (1 + tan0 + sec0) = 2
0 meant theta​

Answer 1

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$(1 + cot \alpha - cosec \alpha )(1 + tan \alpha + sec \alpha ) = 2$

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$change \: tan \: in \: to \: \frac{sin}{cos} or \: cot \: in \: to \: \frac{cos}{sin}$

$= (1 + \frac{cos \alpha }{ sin\alpha } - \frac{1}{ \sin \alpha } )(1 + \frac{ \sin \alpha }{ \cos \alpha } + \frac{1}{cos \alpha } )$

$= ( \frac{sin + cos - 1}{sin} )( \frac{cos + sin + 1}{cos} )$

$using \: suitible \: identity$

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

$= \frac{ {(sin \alpha + cos \alpha) }^{2} - {1}^{2} }{sin \alpha cos \alpha }$

$= \frac{{sin}^{2} \alpha + {cos}^{2} \alpha + 2sin \alpha cos \alpha - 1}{2sin \alpha cos \alpha }$

$using \: identity = {sin}^{2} \alpha + {cos}^{2} \alpha = 1$

$= \frac{1 - 1 + 2sin \alpha cos \alpha }{2sin \alpha \: cos \alpha }$

$= \frac{2sin \alpha \: cos \alpha }{ \sin \alpha \cos \alpha \alpha }$

$= 2$

Answer 2

Step-by-step explanation:

Given : Expression $(1 + \cot \theta - \csc \theta )(1 + \tan \theta + \sec \theta ) = 2$

To find : Prove the expression ?

Solution :

Taking LHS,

Using trigonometric properties,

$\tan\theta=\frac{\sin\theta}{\cos\theta} \ or\ \cot =\frac{cos\theta}{sin\theta}$

$= (1 + \frac{\cos \theta }{ \sin\theta} - \frac{1}{ \sin \theta} )(1 + \frac{ \sin \theta }{ \cos \theta} + \frac{1}{cos \theta} )$

$= ( \frac{\sin \theta+ \cos\theta - 1}{sin} )( \frac{\cos\theta + \sin\theta + 1}{\cos\theta} )$

Using algebraic identity, $(x + y)(x - y) = {x}^{2} - {y}^{2}$

$= \frac{ {(\sin \theta + \cos \theta) }^{2} - {1}^{2} }{\sin \theta \cos\theta } \\\\ = \frac{{\sin\theta}^{2} + {\cos\theta}^{2} + 2\sin\theta \cos\theta- 1}{2\sin \theta \cos\theta }$

Using ${\sin}^{2}\theta+ {\cos}^{2}\theta= 1$

$= \frac{1 - 1 + 2\sin \theta cos \theta }{2\sin \theta \cos\theta}\\\\ = \frac{2\sin \theta cos \theta }{ \sin \theta \cos \theta } \\\\ = 2$

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Prove this trigonometric identities

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