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sin(180-theta)sin(180+theta)cosec^2theta is equal to
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sin (a+b) = sina cos b + sinb cos a sin (a -b) = sina cos b - sinb cos a cosc a=\(\dfrac{1}{sin a}\) if your calculating was right, the answer will be -1
sin180 costheta+sintheta cos180
sin180 cos theta-sin theta cos180
@ooops
The quadratic equation whose roots are sin 18° and cos 36° is
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for the first one sin180 =0, cos 180 =-1
for the second question, it is (x-sin18)(x-cos 36)=0, open the bracket, you have the required quadratic.
x^2-xcos36-xsin18-sin18cos36
you can let it as it is or find out the value of sin 18 and cos 36 from the calculator. I don't have calculator with me, so I can't check.
only by using calculator u can check it.otherwise no formula
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