Prove that:
(1+cot-cosec)(1+tan+sec)=2
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(1+cotA−cosecA)(1+tanA+secA)=2
L.H.S.
=(1+cotA−cosecA)(1+tanA+secA)
=(1+ sinAcosA − sinA1 )
(1+ cosAsinA + cosA1 )
=( sinAsinA+cosA−1)
( cosAcosA+sinA+1 )
= sinA.cosA(sinA+cosA)2−12
= sinA.cosAsin 2 A+cos
2A+2sinA.cosA−1
=
sinA.cosA
1+2sinA.cosA−1
=
sinA.cosA
2sinA.cosA
=2
=R.H.S.
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Answered by
1
Answer:
2
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.
=(sin^2A+cos^2A+2.sinA.cosA-1)/sinA.cosA.
=( 1+2.sinA.cosA -1)/sinA.cosA.
= 2.sinA.cosA/sinA.cosA
= 2 , proved.
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