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# Prove the Following Trigonometric Identities. T_N = Sin^N Theta + Cos^N Theta, Prove that (T_3 - T_5)/T_1 = (T_5 - T_7)/T_3 - CBSE Class 10 - Mathematics

#### Question

Prove the following trigonometric identities.

if T_n = sin^n theta + cos^n theta, prove that (T_3 - T_5)/T_1 = (T_5 - T_7)/T_3

#### Solution

In the given question, we are given T_n = sin^n theta + cos^n theta

We need to prove (T_3 - T_5)/T_1 = (T_5 - T_7)/T_3

Here L.H.S is

(T_3 - T_5)/T_1 = ((sin^3 theta = cos^3 theta) - (sin^5 theta + cos^5 theta))/((sin theta + cos theta))

Now, solving the L.H.S, we get

((sin^3 theta + cos^3 theta)- (sin^5 theta + cos^5 theta))/((sin theta + cos theta)) = (sin^3 theta - sin^5 theta + cos^3 theta - cos^ 5 theta)/(sin theta + cos theta)

 = (sin^3 theta (1 - sin^2 theta) + cos^3 theta (1 - cos^2 theta))/((sin theta  + cos theta))

Further Using the property sin^2 theta + cos^2 theta = 1 we get

cos^2 theta = 1 - sin^2 theta

sin^2 theta = 1 - cos^2 theta

So,

(sin^3 theta(1 - sin^2 theta) + cos^3 theta (1 - cos^2 theta))/(sin theta + cos theta)  = (sin^3 theta cos^2 theta + cos^3 theta sin^2 theta)/(sin theta + cos theta)

= (sin^2 theta cos^2 theta (sin theta + cos theta))/(sin theta + cos theta)

= sin^2 theta cos^2 theta

Now, solving the R.H.S, we get

(T_5 - T_7)/T_3 = ((sin^5 theta + cois ^5)(sin^7 theta + cos^2 theta))/(sin^3 theta + cos^3 theta)

So,

((sin^5 theta + cos^5 theta) - (sin^7 theta + cos^7 theta))/(sin^3 theta + cos^3 theta) = (sin^5 theta - sin^7 theta + cos^5 theta - cos^7 theta)/(sin^3 theta + cos^3 theta)

= (sin^5 theta (1  - sin^2 theta) + cos^5 theta (1 + cos^2 theta))/ (sin^3 theta + cos^3 theta)

= sin^2 theta cos^2 theta

Hence proved

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Solution Prove the Following Trigonometric Identities. T_N = Sin^N Theta + Cos^N Theta, Prove that (T_3 - T_5)/T_1 = (T_5 - T_7)/T_3 Concept: Trigonometric Identities.
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