prove :cos 40⁰ upon sin 50⁰ + sin 20⁰ upon cos 70⁰- tan 75 × tan 15 = 1
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prove :cos 40⁰ upon sin 50⁰ + sin 20⁰ upon cos 70⁰- tan 75 × tan 15 = 1
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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
To Prove:
[tex]\frac{cos40}{sin50}[/tex] + [tex]\frac{sin20} {cos70}[/tex] - [tex][ {tan75}\times{tan15} ][/tex] = 1
Step-by-step explanation:
LHS = [tex]\frac{cos40}{sin50}[/tex] + [tex]\frac{sin20} {cos70}[/tex] - [tex][ {tan75}\times{tan15} ][/tex]
= [tex]\frac{cos40}{sin50}[/tex] + [tex]\frac{sin20} {cos70}[/tex] - [tex]\frac{tan75}{cot15}[/tex]
= [tex]\frac{cos(90-50)}{sin50}[/tex] + [tex]\frac{sin(90-70)} {cos70}[/tex] - [tex]\frac{tan(90-15)}{cot15}[/tex]
= [tex]\frac{sin50}{sin50}[/tex] + [tex]\frac{cos70} {cos70}[/tex] - [tex]\frac{cot15}{cot15}[/tex]
= 1 + 1 - 1
= 2-1
= 1
= RHS
LHS = RHS
hence proved