Question

pls solve it fast trignometry​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Step-by-step explanation:

${\underline{{\text{To Prove:}}}}$

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

${\underline{{\text{Solution:}}}}$

L.H.S

$(1 + \cot A - \cosec A)(1 + \tan A + \sec A) \\ \\ (1 + \frac{ \cos A }{ \sin A} - \frac{1}{ \sin A} )(1 + \frac{ \sin A}{ \cos A } + \frac{1}{ \cos A} ) \\ \\ ( \frac{ \sin A+ \cos A- 1 }{ \sin A} )( \frac{ \cos A + \sin A + 1}{ \cos A} ) \\ \\ ( \frac{ \sin A+ \cos A- 1 }{ \sin} )( \frac{ \sin A+ \cos A + 1 }{ \cos A } ) \\ \\ \frac{( \sin A+ \cos A){}^{2} - {(1)}^{2} }{ \sin A .\cos A} \\ \\ \frac{ { \sin }^{2} A + { \cos {}^{2} A } + 2 \sin A.\cos A - 1}{ \sin A. \cos A} \\ \\ \frac{1 - 1 + 2 \sin A.\cos A }{ \sin A.\cos A} \\ \\ \frac{2 \sin A.\cos A }{ \sin A.\cos A} \\ \\ 2 \\ \\</p><p>\therefore \text{L.H.S = R.H.S [Proved]}$

${{\boxed{ \text{\blue{Some important trigonometry Rules}}}}}$

$sin \theta.cosec\theta = 1 \\ \\</p><p>\cos\theta.\sec\theta = 1 \\ \\</p><p>\tan\theta.\cot\theta = 1\\ \\</p><p>\sin^2\theta + \cos^2 \theta= 1 \\ \\</p><p>\cosec^2 \theta -\cot^2 \theta = 1 \\ \\</p><p>\sec^2 \theta - \tan^2 \theta = 1$