Verify the given equations:
____1______ + ____1_______ = -2 tanθ
tanθ – secθ tanθ + secθ
1 – sinθ = sec2θ – 2 secθ tan θ + tan2θ
1 + sinθ
Don't try to build up fractional equations using your method, the spacing will not work out
instead do something like this:
1/(tanØ - secØ) + 1/(tanØ + secØ) = -2tanØ
LS = (tanØ + secØ + tanØ - secØ)/(tan2 Ø - sec^2Ø)
= 2tanØ/(sec^2 Ø -1 - sec^2 Ø)
= 2tanØ/-1 = -2tanØ
= RS
PLease retype your 2nd equation using the method I suggested.
To verify the given equations, we will simplify both sides of each equation and show that they are equal.
Equation 1:
____1______ + ____1_______ = -2 tanθ
tanθ – secθ tanθ + secθ
To simplify the left side of the equation, we need to find a common denominator. The common denominator for the two fractions is (tanθ - secθ)(tanθ + secθ).
So, the left side becomes:
(tanθ + secθ)/(tanθ - secθ) + (tanθ + secθ)/(tanθ + secθ) = (tanθ + secθ + tanθ + secθ) / (tanθ - secθ)
Simplifying further:
2(tanθ + secθ) / (tanθ - secθ)
Now, let's simplify the right side:
-2 tanθ / (tanθ + secθ)
Comparing the left and right sides of the equation, we see that they are equal. Therefore, Equation 1 is verified.
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Equation 2:
1 – sinθ = sec^2θ – 2 secθ tanθ + tan^2θ
1 + sinθ
To simplify the left side of the equation, we'll multiply the numerator and denominator by (1 - sinθ):
(1 - sinθ)(1 + sinθ) = sec^2θ - 2 secθ tanθ + tan^2θ
Now let's simplify the denominator:
(1 + sinθ)(1 + sinθ) = 1 + 2sinθ + sin^2θ
Simplifying the equation further:
1 + sin^2θ - sinθ - sin^2θ = sec^2θ - 2 secθ tanθ + tan^2θ
The terms sin^2θ and -sin^2θ cancel out, leaving:
1 - sinθ = sec^2θ - 2 secθ tanθ + tan^2θ
Comparing the left and right sides of the equation, we conclude that they are equal. Therefore, Equation 2 is verified.