Question

Find the value of x if,cosec(90°-theta)/sin(90°-theta)- x/tan(90°-theta)=1
Please answer my question...its urgent!!​

Answer 1

$\frac{ \cosec(90 - \theta) }{ \sin(90 - \theta) } - \frac{x}{ \tan(90 -\theta ) } = 1 \\ \\ = > \frac{ \ \sec(\theta) }{ \cos(\theta) } - \frac{x}{ \cot(\theta) } = 1 \\ \\ \small{ \rm{ \bigg\{swicthing \: positions \: \: of \: \: 1 \: \: \& \: \: \frac{x}{\cot(\theta)} \bigg\}}} \\ \\ = > \frac{ \ \sec(\theta) }{ \cos(\theta) } - 1 = \frac{x}{ \cot(\theta)} \\ \small{ \rm{ \{transposing \: \: \cot(\theta) \: \: to \: \: LHS \}}} \\ = > \cot(\theta) \bigg \{\frac{ \ \sec(\theta) }{ \cos(\theta) } - 1 \bigg\} = x \\ \small{ \text{ \{ switching \: \:the \: \: LHS \: \: \& \: \:RHS\}}} \\ = > x = \cot(\theta) \bigg \{ \frac{ \frac{1}{\cos(\theta)} }{\cos(\theta)} - 1 \bigg\} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \because \: \: \sec \theta = \frac{1}{\cos \theta} \\ \\ = > x = \frac{\cos(\theta)}{\sin(\theta)} \bigg \{ \frac{1}{ { \cos}^{2} \theta } - 1 \bigg\}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:\because \: \: \cot \theta = \frac{ \cos \theta}{\sin \theta} \\ \\ \small{ \rm{ \bigg\{ perfoming \: \: multiplictions \bigg\}}} \\ \\ = > x = \frac{1}{\sin(\theta)\cos(\theta)} - \frac{\cos(\theta)}{\sin(\theta)} \\ \small{ \text{ \{ taking \: \: \: \: of \: \: denominators\}}} \\ = > x = \frac{1 - \cos(\theta) \cos(\theta)}{ \sin(\theta)\cos(\theta)} \\ \\ = > x = \frac{1 - { \cos}^{2} \theta }{\sin(\theta)\cos(\theta)} \\ \\ = > x = \frac{{\sin}^{2} \theta}{\sin(\theta)\cos(\theta)} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \{\because \: \:1 - { \cos}^{2} \theta = { \sin}^{2} \theta \}\\ \small{ \rm{ \{ rewriting \: {\sin}^{2} \theta \: as \: \sin(\theta) \times \sin(\theta) \: \}}} \\ = > x = \frac{\sin(\theta)\cancel{\sin(\theta)}}{\cancel{\sin(\theta)}\cos(\theta)} \\ \\ = > x = \frac{ \sin(\theta)}{ \cos(\theta)} \\ \\ = > x = \pink{ \tan( \theta) }$