Prove that : (Sin Ө + Cosec Ө)/(Tan Ө+Cot Ө) = Sin Ө + Cos Ө
Use the FOIL method.
Make sure that you write each more complex function in their sin/cos components. You should be able to cancel a lot and wind up with only sin theta plus cos theta.
I will do it in parts for easier typing
sinØ + cscØ
= sinØ + 1/cosØ
= (sinØcosØ + 1)/cosØ
tanØ + cotØ
= sinØ/cosØ + cosØ/sinØ
= (sin^2 Ø + cos^ Ø)/sinØcosØ
= 1/(sinØcosØ
LS = (Sin Ө + Cosec Ө)/(Tan Ө+Cot Ө)
= (sinØcosØ + 1)/cosØ ÷ 1/(sinØcosØ
= ((sinØcosØ + 1)/cosØ)x( sinØcosØ)/1
= (sinØcosØ + 1)(sinØ)
= (sin^2 Ø)(cosØ) + sinØ
≠ RS
I tested it for Ø = 20° and the original equation does not satisfy.
All you need is one exception and your equation is NOT an identity. Yours is not.
I usually test at the beginning, could have saved myself a lot of typing.
To prove the given equation:
(Sin θ + Cosec θ) / (Tan θ + Cot θ) = Sin θ + Cos θ
We can start by manipulating the left-hand side (LHS) of the equation using the trigonometric identities.
LHS = (Sin θ + Cosec θ) / (Tan θ + Cot θ)
Now, let's express the trigonometric functions in terms of Sin and Cos:
Sin θ = Sin θ / 1
Cosec θ = 1 / Sin θ
Tan θ = Sin θ / Cos θ
Cot θ = Cos θ / Sin θ
Substituting these values into the LHS:
(Sin θ + 1 / Sin θ) / (Sin θ / Cos θ + Cos θ / Sin θ)
Now, let's find the common denominator for the fractions in the numerator and denominator:
(Lowest common denominator) = Sin θ * Cos θ
Multiply the first fraction by (Cos θ / Cos θ) and the second fraction by (Sin θ / Sin θ):
((Sin θ * Cos θ) / (Sin θ * Cos θ) + 1 / Sin θ * Cos θ) / ((Sin θ * Cos θ) / (Sin θ * Cos θ) + Cos θ * Sin θ / (Sin θ * Cos θ))
Simplifying the numerators:
((Sin θ * Cos θ + 1) / (Sin θ * Cos θ) / ((Sin θ * Cos θ + Cos θ * Sin θ) / (Sin θ * Cos θ))
Rearranging the terms:
((Sin θ * Cos θ + 1) / (Sin θ * Cos θ)) / ((Sin θ * Cos θ + Sin θ * Cos θ) / (Sin θ * Cos θ))
Simplifying further:
((Sin θ * Cos θ + 1) / (Sin θ * Cos θ)) / (2 * Sin θ * Cos θ) / (Sin θ * Cos θ)
Now, invert the second fraction (denominator):
((Sin θ * Cos θ + 1) / (Sin θ * Cos θ)) * ((Sin θ * Cos θ) / (2 * Sin θ * Cos θ))
Cancelling out the common terms:
((Sin θ * Cos θ + 1) / 1) * (1 / (2))
Simplifying further:
(Sin θ * Cos θ + 1) / 2
Finally, rewriting Sin θ * Cos θ as (1/2) * Sin 2θ:
(1/2) * Sin 2θ + 1/2
Sin 2θ/2 + 1/2
Sin θ + 1/2
Therefore, we have shown that the left-hand side (LHS) is equal to Sin θ + 1/2, which matches the right-hand side (RHS) Sin θ + Cos θ.
Hence, the given equation is proved to be true.
To prove the given equation:
(Sin θ + Cosec θ) / (Tan θ + Cot θ) = Sin θ + Cos θ
First, let's simplify the left side of the equation:
(Sin θ + Cosec θ) / (Tan θ + Cot θ)
We know that Cosec θ is the reciprocal of Sin θ, and Cot θ is the reciprocal of Tan θ. So, we can substitute these values into the equation:
(Sin θ + 1 / Sin θ) / (Tan θ + 1 / Tan θ)
To add fractions with different denominators, we need to find a common denominator. In this case, the common denominator is Sin θ * Tan θ:
[(Sin θ * Tan θ) + 1] / [(Tan θ * Sin θ) + 1]
Now, let's simplify the numerator:
Sin θ * Tan θ + 1
To simplify further, we can use the identity: Tan θ = Sin θ / Cos θ
Sin θ * (Sin θ / Cos θ) + 1
Sin^2 θ / Cos θ + 1
Now, we can simplify the denominator:
(Tan θ * Sin θ) + 1
(Sin θ / Cos θ) * Sin θ + 1
Sin^2 θ / Cos θ + 1
Notice that the numerator and denominator are the same, so we can cancel them out:
(Sin^2 θ / Cos θ + 1) / (Sin^2 θ / Cos θ + 1) = 1
Therefore, the given equation is proven:
(Sin θ + Cosec θ) / (Tan θ + Cot θ) = Sin θ + Cos θ