Question

show
((sec2θ-1)/(2sec2θsinθ))=sinθ

Answer 1

anyone can guide me for the first step?

Answer 2

is it sec^2 θ or sec2θ?

Answer 3

no power

Answer 4

@TuringTest @UnkleRhaukus

Answer 5

\[\frac{\sec(2\theta)-1}{2\sec(2\theta)\sin(\theta)}\]
\[=\frac{\frac1{\cos(2\theta)}-1}{2\frac1{\cos(2\theta)}\sin(\theta)}\]
now multiply terms in the numerator And in the denominator by \(\cos(2\theta)\)

and then use a half angle formula in the numerator

Answer 6

Any problem you are facing @mathnoob1

Answer 7

i having problem seeing the next step after UnkleRhaukus 1st step

Answer 8

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

Answer 9

\[=\frac{\frac1{\cos(2\theta)}-1}{2\frac1{\cos(2\theta)}\sin(\theta)} \times \frac{ \cos (2 \theta) }{ \cos (2 \theta) }\]\[=\frac{1-\cos(2 \theta)}{2\sin(\theta)}\]\[\frac{ 1-\cos(2 \theta) }{ 2 }=\sin^2\theta\]You can finish from here

Answer 10

hmm i still lost from the 2nd step to the 3rd step..

Answer 11

\[=\frac{1-\cos(2 \theta)}{2\sin(\theta)}\]
\[=\frac{1-\cos(2 \theta)}{2}\times\frac1{\sin(\theta)}\]



this is the half angle formula \[\boxed{\frac{ 1-\cos(2 \theta) }{ 2 }=\sin^2\theta}\]