prove that:
(cosecA-sinA)(secA-cosA)*sec²A=tanA
Answers
Answered by
14
CosecA-sinA=1-sin^2A/sinA.
and, secA-cosA=1-cos^2A/cosA
now, putting on ur question.from lhs.
1-sin^2A/sinA*1-cos^2A/cosA*sec^2A
=) where ,,1 -sin^2=cos^2A.
and ,1-cos^2A=sin^2A.
so, cos^2A/sinA*sin^2A/cosA*sec^2A
=)cos^2A/sinA*Sin^2A/cosA*1/cos^2A 【cos^2Acancelled】
=)remaining,
sin^2A/sinA*cosA
=)sinA/cosA
=)tanA Rhs,
prooved ...
hope it help you..
@rajukumar☺☺1
and, secA-cosA=1-cos^2A/cosA
now, putting on ur question.from lhs.
1-sin^2A/sinA*1-cos^2A/cosA*sec^2A
=) where ,,1 -sin^2=cos^2A.
and ,1-cos^2A=sin^2A.
so, cos^2A/sinA*sin^2A/cosA*sec^2A
=)cos^2A/sinA*Sin^2A/cosA*1/cos^2A 【cos^2Acancelled】
=)remaining,
sin^2A/sinA*cosA
=)sinA/cosA
=)tanA Rhs,
prooved ...
hope it help you..
@rajukumar☺☺1
Answered by
0
Step-by-step explanation:
(1/sinA-sinA) (1/cosA-cosA) sec²A
(1-sin²A/sinA) (1-cos²A/cosA) sec²A
(cos²A/sinA) (sin²A/cosA) sec²A
sinA/cosA=tanA=RHS
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