Question

sin4q-cos4q=sin2q-cos2q​

Answer 1

Answer:

sin⁴Q-cos⁴Q = sin²Q-cos²Q

Step-by-step explanation:

LHS = sin⁴Q-cos⁴Q

= (Sin²Q)² - (cos²Q)²

= (sin²Q+cos²Q)(sin²Q-cos²Q)

/* By algebraic identity:

a²-b² = (a+b)(a-b) */

= 1(sin²Q-cos²Q)

/* By Trigonometric identity:

Sin²A+cos²A = 1 */

= sin²Q-cos²Q

= RHS

Therefore,

sin⁴Q-cos⁴Q = sin²Q-cos²Q

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Answer 2

Given:

$\sin^4 q- \cos^4 q = \sin^2q-\cos^2 q$

To prove:

L.H.S =R.H.S

Solution:

$\sin^4 q- \cos^4 q = \sin^2q-\cos^2 q$

Formula:

$a^2-b^2= (a+b)(a-b)$

Solve L.H.S part:

$\Rightarrow\sin^4 q- \cos^4 q \\\\ \Rightarrow (\sin^2 q)^2 -(\cos^2 q)^2 \\\\ \Rightarrow(\sin^2 q+\cos^2 q) (\sin^2 q-\cos^2 q) \\\\\therefore \sin^2 q+\cos^2 q =1\\\\ \Rightarrow 1 \times (\sin^2 q-\cos^2 q)\\\\ \Rightarrow \sin^2 q-\cos^2 q$

The final value is L.H.S=R.H.S