Question

please solve this question for me

Answer 1

LHS
(sinA*sinA/cosA)/1-cosA
sin^2A/1-cos^2A
sin^2/sin^2
=1

RHS
1+1/cosA
cosA+1/cosA

Answer 2

◢GIVEN

$= > \frac{sina.tana}{1 - cosa} = 1 + seca$
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● WE GIVE LHS

$= > \frac{sina.tana}{1 - cosa} \\ \\ = > \frac{sina.tana}{1 - cosa} \times \frac{1 + cosa}{1 + cosa} \\ \\ = > \frac{(sina.tana) \times (1 + cosa)}{(1 - {cos}^{2}a) }$
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● USING SOME TRIGONOMETRIC PROPERTIES :-

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

◢AND

$= > tan \alpha = \frac{sin \alpha }{cos \alpha }$
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◢THEN

$= > \frac{(sina.tana)(1 + cosa)}{1 - {cos}^{2} a} \\ \\ = > \frac{(sina \times \frac{sina}{cosa} )(1 + cosa)}{ {sin}^{2}a } \\ \\ = > \frac{ {sin}^{2}a(1 + cosa) }{ {sin}^{2}a.cosa } \\ \\ = > \frac{1 + cosa}{cosa} \\ \\ = > \frac{1}{cosa} + \frac{cosa}{cosa} \\ \\ = > seca + 1$
_____________________[◢PROVED]

◢HENCE

=> R. H. S. = L. H. S

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☆$BE \: \: BRAINLY$☆