Question

prove that : -1 +sinA sin(90 - A) / cot(90 - A) = -sin^2 A​

Answer 1

Step-by-step explanation:

this how you do this question

Answer 2

$\huge \red{ \mathfrak{solution}}$

To prove :

$\boxed{ \bold{ - 1 + \frac{ \: sin \theta \: sin(90 - \theta)}{cot \: (90 - \theta)} = - {sin}^{2} \theta}}$

LHS

$\star \: - 1 + \frac{sin \theta \: sin \: (90 - \theta)}{cot \: (90 - \theta)} \\ \\ \star \: - 1 + \frac{sin \theta \: cos \theta}{tan \theta} \\ \\ \star \: - 1 + \frac{sin \theta \ \: cos \theta }{ \frac{sin \theta}{cos \theta} } \\ \\ \star \: - 1 + \frac{sin \theta \: {cos}^{2} \theta }{sin \theta} \\ \\ \star \: - 1 + \frac{ \cancel{sin \theta} \: cos ^{2} \theta}{ \cancel{ sin \theta}} \\ \\ \star \: - 1 + {cos}^{2} \theta \\ \\ \star \: \: - {sin}^{2} \theta$

LHS = RHS

$\huge\boxed{ \blue{ \mathfrak{proved}}}$

Formula used :

$\rightarrow sin(90 - \alpha ) = \cos( \alpha ) \\ \\ \rightarrow \: \cot(90 - \alpha ) = \tan( \alpha ) \\ \\ \rightarrow \: { \sin}^{2} \alpha + { \cos}^{2} \alpha = 1$