Question

sec^2(A) + cosec^2(A) = sec^2(A).cosec^2(A)
@michele_liano

Answer 1

@Michele_Laino

Answer 2

we have to apply these identities:
\[\sec A = \frac{1}{{\cos A}},\quad \csc A = \frac{1}{{\sin A}}\]

Answer 3

?

Answer 4

have u tried to solve the expression?? try converting the terms into sin nd cos form nd then take the LCM nd solve

Answer 5

left side becomes:
\[\sec A + {\left( {\csc A} \right)^2} = \frac{1}{{\cos A}} + {\left( {\frac{1}{{\sin A}}} \right)^2} = ...\]
please continue

Answer 6

i'm confused

Answer 7

I think that there is a typo into your original expression, please check

Answer 8

fixed

Answer 9

ok! so left side becomes:
\[{\left( {\sec A} \right)^2} + {\left( {\csc A} \right)^2} = {\left( {\frac{1}{{\cos A}}} \right)^2} + {\left( {\frac{1}{{\sin A}}} \right)^2} = ...\]

Answer 10

ok m calling sinA = x
nd cosA=y

1/y^2 +1/x^2
take LCM we get -

(x^2 + y^2)/(xy)^2

we know that sin^2A+cos^2A = 1
therefore the numerator = 1
nd we r left with

1/(xy)^2
=sec^A cosec^A

Answer 11

sin^2(A)+cos^2(A)/cos(A)sin(A) = RHS

Answer 12

yes

Answer 13

1/sin(A)cos(A) = sec(A)cosec(A) = RHS

Answer 14

more precisely, it is:
sin^2(A)+cos^2(A)/cos(A)^2 sin(A)^2 = RHS

Answer 15

ok!thanks...

Answer 16

i nedd more help
*need