prove that (sec A+ tan A-1) (sec A-tan A+1)= 2tanA
Answers
Answered by
238
[secA+(tanA-1)(secA-(tanA-1)]
=sec²A-(tanA-1)²
=1/cos²A-(tan²A-2tanA+1)
=1+tan²A-tan²A+2tanA-1
=(1-1)tan²A-tan²A + 2tanA
=0+0+2tanA
=2tanA=2tanA
hence verified!
=sec²A-(tanA-1)²
=1/cos²A-(tan²A-2tanA+1)
=1+tan²A-tan²A+2tanA-1
=(1-1)tan²A-tan²A + 2tanA
=0+0+2tanA
=2tanA=2tanA
hence verified!
Answered by
65
Answer:
Sec^2 a - sec a . tan a + tan a . sec a - tan^2 a + tan a - sec a + tan a - 1
=Sec^2 a - tan^2 a + 2tan a - 1
= 1 + 2tan a - 1
=2tan a...
Hence proved
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