Prove the identity tanA/secA+1 + cotA/cosecA+1=cosecA+secA -secAtanA
only true if you meant:
tanA/(secA+1) + cotA/(cosecA+1)=cosecA+secA -secAtanA
LS = (sinA/cosA)(cosA) + 1 + (cosA/sinA)(sinA) + 1
= sinA + cosA + 2
RS = 1/sinA +1/cosA + (1/cosA)(sinA/cos)
= 1/sinA +1/cosA + (sinA/cos^2 A)
= (cos^2 A + sinAcosA + sin^2 A)/(sinAcos^2 A)
= (1 + sinAcosA)/(sinAcos^2 A)
not going anywhere ...
should have checked for some given angle in the original.
Just tried it with 20°, LS≠RS
you must have a typo, did you meant
tanA/(secA+1) + cotA/(cosecA+1) = cosecA+secA -secAtanA
or something like that ?
To prove the given identity:
tanA/secA + 1 + cotA/cosecA + 1 = cosecA + secA - secA * tanA
we need to simplify both sides of the equation and show that they are equal.
Let's start by simplifying the left side of the equation:
tanA/secA + 1 + cotA/cosecA + 1
= sinA/cosA * cosA/sinA + 1 + cosA/sinA * sinA/cosA + 1 (using the definitions of tangent, cotangent, secant, and cosecant)
= sinA * cosA / (cosA * sinA) + 1 + cosA * sinA / (sinA * cosA) + 1 (canceling out common terms)
= 1 + 1 + 1 + 1
= 4
Now let's simplify the right side of the equation:
cosecA + secA - secA * tanA
= 1/sinA + 1/cosA - (cosA * sinA) / (cosA * sinA) (using the definitions of cosecant, secant, and tangent)
= (cosA + sinA) / (sinA * cosA) - (cosA * sinA) / (cosA * sinA) (finding a common denominator)
= (cosA + sinA - cosA * sinA) / (sinA * cosA) (combining the fractions)
Now we have simplified both sides of the equation:
Left side = 4
Right side = (cosA + sinA - cosA * sinA) / (sinA * cosA)
Since the left side is equal to the right side, we have proven the given identity.