Question

please mark these answers fast friends please friends​

Answer 1

Answer:

As we know that, sin(90°-θ) = cos θ by sine property. So, sin(45°+θ)-cos(45°-θ)  = sin[90°-(45°-θ)]-cos(45°-θ) = cos(45°-θ)-cos(45°-θ) (By using identity)

Answer 2

Step-by-step explanation:

$\rm tan30° = \big( \frac{tan60° - tan30°}{1 + tan60°tan30°} \big)$

$\rm = \frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \cancel{\sqrt{3}} \times \frac{1}{ \cancel{ \sqrt{3} } } } \: \{ \because \: tan60 = \sqrt{3} \: and \: tan30 = \frac{1}{ \sqrt{3} } \}$

$= \frac{ \frac{({ \sqrt{3} })^{2} - 1}{ \sqrt{3} } }{ 1 + 1 }$

$= \frac{ \frac{3 - 1}{ \sqrt{3} } }{2}$

$= \frac{ \frac{2}{ \sqrt{3} } }{2}$

$= \frac{ \cancel2}{ \sqrt{3} } \times \frac{1}{\cancel2}$

$= \frac{1}{ \sqrt{3} }$

$= \tan30° \: \: \{ \because \: \frac{1}{ \sqrt{3} } = \tan30 ° \}$

I hope it is helpful