Prove the following identity:-
(i) tan^3x/1+tan^2x + cot^3x/1+cot^2 = 1-2sin^x cos^x/sinx cosx
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To prove the given identity, we will start with the left-hand side (LHS) and simplify it until it matches the right-hand side (RHS) of the equation.
Given identity: (tan^3x)/(1+tan^2x) + (cot^3x)/(1+cot^2x) = (1-2sin^x cos^x)/(sinx cosx)
Step 1: Express the tangents and cotangents in terms of sine and cosine.
Recall that tan(x) = sin(x)/cos(x) and cot(x) = cos(x)/sin(x).
Using these identities, we can rewrite the LHS as follows:
(tan^3x)/(1+tan^2x) + (cot^3x)/(1+cot^2x)
= (sin^3x/cos^3x)/(1+sin^2x/cos^2x) + (cos^3x/sin^3x)/(1+cos^2x/sin^2x)
Step 2: Simplify the denominators.
For the first term, we can simplify the denominator using the identity: a^2 - b^2 = (a + b)(a - b).
Applying this identity to the denominator (1+sin^2x/cos^2x), we get:
1 + sin^2x/cos^2x = (cos^2x + sin^2x)/cos^2x
= 1/cos^2x
Similarly, for the second term, we have:
(1+cos^2x/sin^2x) = 1/sin^2x
Now, let's substitute these simplified denominators into the expression:
(sin^3x/cos^3x)/(1/cos^2x) + (cos^3x/sin^3x)/(1/sin^2x)
= (sin^3x/cos^3x)*(cos^2x) + (cos^3x/sin^3x)*(sin^2x)
= sin^x cos^2x + cos^x sin^2x
Step 3: Simplify and factor out common terms.
sin^x cos^2x + cos^x sin^2x
= sin^x cos^x (cos^x + sin^x)
Step 4: Rewrite the right-hand side of the equation.
Recall that sin 2x = 2sin x cos x.
Using this identity, we can rewrite the RHS as:
(1-2sin^x cos^x)/(sinx cosx)
= 1/(sinx cosx) - 2sin^x cos^x/(sinx cosx)
= cscx secx - 2sin^x cos^x/(sinx cosx)
= 1/sinx * 1/cosx - 2sin^x cos^x/(sinx cosx)
Step 5: Simplify the right-hand side.
1/sinx * 1/cosx - 2sin^x cos^x/(sinx cosx)
= (1/sinx)*(1/cosx) - 2(sin^x cos^x)/(sinx cosx)
= cotx tanx - 2(sin^x cos^x)/(sinx cosx)
= cotx tanx - 2sin x cos x
Step 6: Conclusion
We have shown that the LHS is equal to the RHS. Therefore, the given identity is proven:
(tan^3x)/(1+tan^2x) + (cot^3x)/(1+cot^2x) = (1-2sin^x cos^x)/(sinx cosx)