(Sec ∅+tan∅) (1-sin ∅) =cos ∅
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Answered by
2
(Sec A + tan A)(1- sinA)
=(1/cos A +sin A/cosA) (1-sinA)
= (1+sinA)(1-sinA) /cosA
=(1-sin^2 A )/cos A
= cos^2 A /cos A
= cos A.
=(1/cos A +sin A/cosA) (1-sinA)
= (1+sinA)(1-sinA) /cosA
=(1-sin^2 A )/cos A
= cos^2 A /cos A
= cos A.
Answered by
2
Hi there !
LHS :-
(Sec∅ + tan ∅)(1- sin∅)
(1/cos ∅ +sin∅/cos∅) (1-sin∅)
(1+sin∅)(1-sin∅) /cos∅
[ (a+b)(a -b) = a² - b² ]
(1-sin² ∅ ) /cos ∅
[ 1-sin² ∅ = cos² ∅ ]
= cos² ∅ /cos ∅
= cos ∅.
= RHS
LHS :-
(Sec∅ + tan ∅)(1- sin∅)
(1/cos ∅ +sin∅/cos∅) (1-sin∅)
(1+sin∅)(1-sin∅) /cos∅
[ (a+b)(a -b) = a² - b² ]
(1-sin² ∅ ) /cos ∅
[ 1-sin² ∅ = cos² ∅ ]
= cos² ∅ /cos ∅
= cos ∅.
= RHS
manjeetsaini:
Nyc,,, ty for help
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