Question

The value of (cos cube 20-cos cube70)/ (sin cube 70-sin cube 20)

Answer 1

QUESTION :

(cos³20-cos³70)/(sin³70-sin³20)

ANSWER :

now

as we know that

= (cos³20-cos³70)/(sin³70-sin³20)

so = ((cos20)³-(cos70)³)/((sin70)³-(sin20)³)

now as we know that

sin(90-A)=cosA

also

cos(90-A)=sinA

so,

by putting the above in equation

= ((cos(90-70))³-(cos70)³)/((sin70)³-(sin90-70)³)

= ((sin70)³-(cos70)³)/((sin70)³-(cos70)³)

= (sin³70-cos³70)/(sin³70-cos³70)

= 1

as the numerator and denominator are same we can cancel both

so the value of (cos³20-cos³70)/(sin³70-sin³20)
is 1

Answer 2

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