prove that (1+ cot a - cose a ) ( 1+ tan a + sec a)=2
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Answered by
0
Answer:
Step-by-step explanation:
(1+cot A-cosec A).(1+tanA+secA)= 2
L.H.S. =(1+cosA/sinA-1/sinA).(1+sinA/cosA+1/cosA)
=(sinA+cosA-1)×(cosA+sinA+1)/sinA.cosA
=[(sinA+cosA)² -(1)²] / sinA.cosA.
=(sin²A+cos²A+2.sinA.cosA - 1) / sinA.cosA. ( sin²A+cos²A= 1 )
= ( 1 + 2.sinA.cosA -1) / sinA.cosA.
= 2.sinA.cosA / sinA.cosA
= 2
= R.H.S.
Answered by
1
Answer:
Step-by-step explanation:
(1+ cot a - cosec a ) ( 1+ tan a + sec a) = 2
(1+cos a/ sin a - 1/sin a ) (1+ sin a/cos a+1/cos a)
(sin a+cos a - 1 )/sin a* (cos a+sin a+1)/cos a
(sin a*cos a+sin² a+sin a+cos² a+cos a*sin a+cosa-cos a-sin a-1)/sina*cosa
(1-1+2 sin a cos a)/sin a*cos a
2 sin a*cos a/sina cos a
2
hence proved
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