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        1+sec(theta)-tan(theta)/ 1+sec(theta)+tan(theta) = and given options are 	1-cos(theta)/sin(theta)	1-sin(theta)/cos(theta)	1+cos(theta)/sin(theta)	1+sin(theta)/cos(theta)
2 years ago

Lab Bhattacharjee
121 Points
										$\sec^2y-\tan^2y=1 \implies\dfrac{1+\sec y-\tan y}{1+\sec y+\tan y} =\dfrac{\sec^2y-\tan^2y+\sec y-\tan y}{1+\sec y+\tan y} =\sec y-\tan y=\dfrac{1-\sin y}{\cos y}$

2 years ago
Kelvin
11 Points
										Stupid no clear explanations Dont ever give shprt answers And make a fool like u even foolersWorst fellows...

16 days ago
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