(1/sec2A-cos2A + 1/cosec2A-sin2A) (sin2Acos2A) = 1-sin2Acos2A/2+sin2Acos2A - Maths - Some Applications of Trigonometry

Question

(1/sec2A-cos2A +
1/cosec2A-sin2A)
(sin2Acos2A) =
1-sin2Acos2A/2+sin2Acos2A

Priyamtejpal · Answer

L H S
= [1/(sec2A - cos2A) +
1/(cosec2A - sin2A)
]sin2Acos2A
= [1/(1/cos2A - cos2A) +
1/(1/sin2A - sin2A)
]sin2Acos2A
= [cos2A/(1 - cos4A) +
sin2A/(1 - sin4A)
]sin2Acos2A
= [cos2A(1 - sin4A)/(1 -
cos4A)(1 - sin4A) + sin2A(1 -
cos4A)/(1 - sin4A)(1 - cos4A)
]sin2Acos2A
= [sin2A(1 - cos4A) +
cos2A(1 - sin4A)
]sin2Acos2A/ (1 - sin4A)(1 -
cos4A)
= [sin2A -
sin2Acos4A + cos2A –
sin4Acos2A] sin2Acos2A/
(1 - sin4A)(1 -
cos4A)
= [sin2A -
sin2Acos4A + cos2A –
sin4Acos2A] sin2Acos2A
/(1 - sin4A)(1 -
cos4A)
= [sin2A + cos2A -
sin2Acos2A(sin2A +
cos2A) ]sin2Acos2A/ (1 -
sin4A)(1 - cos4A)
= [1 -
sin2Acos2A(1)]sin2Acos2A/
[1 + sin4Acos4A - sin4A -
cos4A]
= [1 -
sin2Acos2A]sin2Acos2A/
[1 + sin4Acos4A – (sin2A +
cos2A)2 +
2sin2Acos2A]
= [1 -
sin2Acos2A]sin2Acos2A/
[1 + sin4Acos4A – (1)2 +
2sin2Acos2A]
= [1 -
sin2Acos2A]sin2Acos2A/
[sin2Acos2A(sin2Acos2A
+ 2)]
= [1 - sin2Acos2A]/
[sin2Acos2A + 2]
= R H S
Hence proved