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\sqrt{1+cosA/1-cosA}
\sqrt{(1+cosA/1-cosA)*(1+cosA/1+cosA)}
\sqrt{(1+cosA)^2/(1-cos^2A)}
(1+cosA)/sinA; [\sqrt{1-cos^2A}=sinA]
(1/sinA)+(cosA/sinA);
cosecA+cotA
proved.......
yes!
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samson70:
i am bot understanding but
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hey
your question answer is.
we have to prove LHS equal to RHS.
so first we take RHS.
that is.
cosec A+cot A
1/sin A+cos A/sin A
1+cos A/sin A .....1eq
now taking LHS.
that is.
✓1+cos A/✓1-cos A
rationalise it
we get
✓1+cos A/✓1-cos A×✓1+cos A/✓1+cos A
✓1+cos A×✓1+cos A/✓1-cos A×✓1+cos A
✓(1+cos A)^2
__________. we know
✓(1-cos A)(1+cos A). (a+b)(a-b)=a^2-b^2...8eq
1+cos A. as root cancel from square
__________
✓1^2-cos A^2.by 8eq 1 ^2-cosA^2=sinA^2
.......6eq
1+cos A
______
✓sin^2A. by ..6eq
1+cos A
_____
sin A. as root cancelled from square
1+cos A
______.......2eq
sin A
hence
1eq=2eq
LHS=RHS
hence proved
___________
×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×
your question answer is.
we have to prove LHS equal to RHS.
so first we take RHS.
that is.
cosec A+cot A
1/sin A+cos A/sin A
1+cos A/sin A .....1eq
now taking LHS.
that is.
✓1+cos A/✓1-cos A
rationalise it
we get
✓1+cos A/✓1-cos A×✓1+cos A/✓1+cos A
✓1+cos A×✓1+cos A/✓1-cos A×✓1+cos A
✓(1+cos A)^2
__________. we know
✓(1-cos A)(1+cos A). (a+b)(a-b)=a^2-b^2...8eq
1+cos A. as root cancel from square
__________
✓1^2-cos A^2.by 8eq 1 ^2-cosA^2=sinA^2
.......6eq
1+cos A
______
✓sin^2A. by ..6eq
1+cos A
_____
sin A. as root cancelled from square
1+cos A
______.......2eq
sin A
hence
1eq=2eq
LHS=RHS
hence proved
___________
×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×-×
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