Question

prove :cos 40⁰ upon sin 50⁰ + sin 20⁰ upon cos 70⁰- tan 75 × tan 15 = 1

Answer 1

Step-by-step explanation:

we know that cos(90-A)=sinA

and sin(90-A)=cosA

and tan(90-A)=cotA

hence cos40°\sin50°+sin20°\cos70°-tan75°×tan15°

=cos40°\cos40°+sin20°\sin20°-tan75°cot75°

=1+1-1

=1

Answer 2

To Prove:

$\frac{cos40}{sin50}$ + $\frac{sin20} {cos70}$ - $[ {tan75}\times{tan15} ]$  = 1

Step-by-step explanation:

LHS = $\frac{cos40}{sin50}$ + $\frac{sin20} {cos70}$ - $[ {tan75}\times{tan15} ]$

= $\frac{cos40}{sin50}$ + $\frac{sin20} {cos70}$ - $\frac{tan75}{cot15}$

= $\frac{cos(90-50)}{sin50}$ + $\frac{sin(90-70)} {cos70}$ - $\frac{tan(90-15)}{cot15}$

= $\frac{sin50}{sin50}$ + $\frac{cos70} {cos70}$ - $\frac{cot15}{cot15}$

= 1 + 1 - 1

= 2-1

= 1

= RHS

LHS = RHS

hence proved