Question

sinѲ+cosѲ/sinѲ-cosѲ + sinѲ-cosѲ/sinѲ+cosѲ= 2/2sin²Ѳ-1
please help me ​

Answer 1

Answer:

sinѲ = 1 / cscѲ cosѲ = 1 / secѲ tanѲ = 1 / cotѲ ... cos (2Ѳ) = 1 - 2sin²Ѳ. double angle formula for tan ... tan (Ѳ/2) = ±√((1-cosѲ)/(1+cosѲ))

Step-by-step explanation:

Hope it's helpful for you

Answer 2

$LHS = \frac{(sin \theta + cos \theta)}{(sin \theta - cos \theta )} + \frac{(sin \theta - cos \theta)}{(sin \theta + cos \theta )} \\= \frac{(sin\theta + cos \theta )^{2} + (sin \theta - cos \theta)^{2}}{(sin \theta - cos \theta )(sin \theta + cos \theta)} \\= \frac{2(sin^{2} \theta + cos^{2}\theta)}{sin^{2} \theta - cos^{2} \theta } \\= \frac{ 2 \times 1 }{ sin^{2}\theta - (1 - sin^{2} \theta )}\\= \frac{2}{sin^{2} \theta - 1 + sin^{2} \theta }\\= \frac{2}{2sin^{2} \theta - 1 } \\= RHS$

•••♪