Asked by Chay
Claim: For all theta such that
-pie/2<theta<pie/2 the following holds true:
(1+tan(theta))^2=1/cos(theta)
-pie/2<theta<pie/2 the following holds true:
(1+tan(theta))^2=1/cos(theta)
Answers
Answered by
Reiny
LS = (1+ tanØ)^2
= 1 + 2tanØ + tan^2 Ø
= 1 + 2tanØ + sec^2 Ø - 1
= 2sinØ/cosØ + 1/cos^2 Ø
= (2sinØcosØ + 1)/cos^2 Ø
= (sin^2 Ø + cos^2 Ø + 2sinØcosØ)/cos^2 Ø
= (sinØ + cosØ)^2 /cosØ
≠ RS
The identity is false , all we need is one example when it does not work
e.g. Ø=30°
LS = (1+tan30°)^2 = appr 2.488
RS = 1/cos30° = 1.15
= 1 + 2tanØ + tan^2 Ø
= 1 + 2tanØ + sec^2 Ø - 1
= 2sinØ/cosØ + 1/cos^2 Ø
= (2sinØcosØ + 1)/cos^2 Ø
= (sin^2 Ø + cos^2 Ø + 2sinØcosØ)/cos^2 Ø
= (sinØ + cosØ)^2 /cosØ
≠ RS
The identity is false , all we need is one example when it does not work
e.g. Ø=30°
LS = (1+tan30°)^2 = appr 2.488
RS = 1/cos30° = 1.15
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