1 /cosec 0 (1 - coto)+ 1/
sec 0 (1 -tan 0) is equal to
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Prove that:(1+cotA−cosecA)(1+tanA+secA)=2.
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LHS =1+cotA−cosecA=1+
sinA
cosA
−
sinA
1
=
sinA
(sinA+cosA−1)
1+tanA+secA=1+
cosA
sinA
+
cosA
1
=
cosA
(sinA+cosA+1)
(1+cotA−cosecA)(1+tanA+secA)
=
sinAcosA
(sinA+cosA−1)(sinA+cosA+1)
=
sinAcosA
(sinA+cosA)
2
−1
=
sinAcosA
1+2sinAcosA−1
=2=RHS
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