Please solve the question 6,7
Q.(vi) $\sin \left(60°-\theta \right)=\cos (30°+\theta )$
[Hint. LHS = $\sin \left[90°-\left(30°+\theta \right)\right]$]
(vii) $\frac{\sin \theta }{\sin \left(90°-\theta \right)}+\frac{\cos \theta }{\cos \left(90°-\theta \right)}$

Dear Student,

$$\begin{gathered} \left(vi\right) To prove: \sin (60° - \theta ) = \cos (30° + \theta ) \\ Proof: LHS = \sin (60° - \theta ) = \cos (90° - (60° - \theta \left)\right) = \cos (90° - 60° + \theta ) = \cos (30° + \theta ) = RHS. Hence, proved. [\because \cos (90° - \theta ) = \sin \theta ] \\ \left(vii\right) \frac{\sin \theta }{\sin (90° - \theta )} + \frac{\cos \theta }{\cos (90° - \theta )} \\ = \frac{\sin \theta }{\cos \theta } + \frac{\cos \theta }{\sin \theta } [\because \sin (90° - \theta ) = \cos \theta , \cos (90° - \theta ) = \sin \theta ] \\ = \frac{{\sin }^2\theta + {\cos }^2\theta }{\sin \theta \cos \theta } \\ = \frac{1}{\sin \theta \cos \theta } [\because {\sin }^2\theta + {\cos }^2\theta = 1] \\ = \cos ec\theta sec\theta \\ = sec(90° - \theta )\cos ec(90° - \theta ) [\because sec(90° - \theta ) = \cos ec\theta , \cos ec(90° - \theta ) = sec\theta ] \end{gathered}$$

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