Question

$\bold \red{question}$


prove that✔️​

cos^4A -cos^2A= sin^4A - sin ^2 A



wrong Answer ⚠️ Will be reported​

Answer 1

Required Answer:-

Given to prove:

$\rm \mapsto \cos^{4} (x) - { \cos}^{2} (x) = { \sin}^{4} (x) - { \sin}^{2} (x)$

Proof:

Taking LHS,

$\rm { \cos}^{4} (x) - { \cos}^{2} (x)$

$\rm = { \cos}^{2} (x) \{ { \cos}^{2} (x) - 1 \}$

$\rm = { \cos}^{2} (x) \times - 1 \{ 1 - { \cos}^{2} (x)\}$

$\rm = - { \cos}^{2} (x) \{ 1 - { \cos}^{2} (x)\}$

We know that,

$\rm \implies { \sin}^{2} (x) + { \cos }^{2} (x) = 1$

$\rm \implies { \sin}^{2} (x) = 1 - { \cos }^{2} (x)$

So,

$\rm- { \cos}^{2} (x) \{ 1 - { \cos}^{2} (x)\}$

$\rm = - { \cos}^{2} (x) \times \sin ^{2} (x)$

$\rm = { \cos}^{2} (x) \times (- \sin ^{2} (x) )$

As we know that,

$\rm \implies { \cos }^{2} (x) = 1 - { \sin }^{2} (x)$

Therefore,

$\rm { \cos}^{2} (x) \times (- \sin ^{2} (x) )$

$\rm = (1 - \sin^{2} (x)) \times (- \sin ^{2} (x) )$

Now, multiply,

$\rm = - \sin ^{2} (x) + { \sin }^{4} (x)$

$\rm = { \sin }^{4} (x) - \sin ^{2} (x)$

Taking RHS,

$\rm = { \sin }^{4} (x) - \sin ^{2} (x)$

Hence, LHS = RHS (Hence Proved)

Important Formula:

• $\rm { \sin}^{2}(x) + { \cos}^{2} (x) = 1$
• $\rm { \csc}^{2} (x) - \cot^{2} (x) = 1$
• $\rm { \sec}^{2} (x) - \tan^{2} (x) = 1$

Answer 2

Answer:

Required Answer:-

Given to prove:

\rm \mapsto \cos^{4} (x) - { \cos}^{2} (x) = { \sin}^{4} (x) - { \sin}^{2} (x)↦cos

4

(x)−cos

2

(x)=sin

4

(x)−sin

2

(x)

Proof:

Taking LHS,

\rm { \cos}^{4} (x) - { \cos}^{2} (x)cos

4

(x)−cos

2

(x)

\rm = { \cos}^{2} (x) \{ { \cos}^{2} (x) - 1 \}=cos

2

(x){cos

2

(x)−1}

\rm = { \cos}^{2} (x) \times - 1 \{ 1 - { \cos}^{2} (x)\}=cos

2

(x)×−1{1−cos

2

(x)}

\rm = - { \cos}^{2} (x) \{ 1 - { \cos}^{2} (x)\}=−cos

2

(x){1−cos

2

(x)}

We know that,

\rm \implies { \sin}^{2} (x) + { \cos }^{2} (x) = 1⟹sin

2

(x)+cos

2

(x)=1

\rm \implies { \sin}^{2} (x) = 1 - { \cos }^{2} (x)⟹sin

2

(x)=1−cos

2

(x)

So,

\rm- { \cos}^{2} (x) \{ 1 - { \cos}^{2} (x)\}−cos

2

(x){1−cos

2

(x)}

\rm = - { \cos}^{2} (x) \times \sin ^{2} (x)=−cos

2

(x)×sin

2

(x)

\rm = { \cos}^{2} (x) \times (- \sin ^{2} (x) )=cos

2

(x)×(−sin

2

(x))

As we know that,

\rm \implies { \cos }^{2} (x) = 1 - { \sin }^{2} (x)⟹cos

2

(x)=1−sin

2

(x)

Therefore,

\rm { \cos}^{2} (x) \times (- \sin ^{2} (x) )cos

2

(x)×(−sin

2

(x))

\rm = (1 - \sin^{2} (x)) \times (- \sin ^{2} (x) )=(1−sin

2

(x))×(−sin

2

(x))

Now, multiply,

\rm = - \sin ^{2} (x) + { \sin }^{4} (x)=−sin

2

(x)+sin

4

(x)

\rm = { \sin }^{4} (x) - \sin ^{2} (x)=sin

4

(x)−sin

2

(x)

Taking RHS,

\rm = { \sin }^{4} (x) - \sin ^{2} (x)=sin

4

(x)−sin

2

(x)

Hence, LHS = RHS (Hence Proved)

Important Formula:

\rm { \sin}^{2}(x) + { \cos}^{2} (x) = 1sin

2

(x)+cos

2

(x)=1

\rm { \csc}^{2} (x) - \cot^{2} (x) = 1csc

2

(x)−cot

2

(x)=1

\rm { \sec}^{2} (x) - \tan^{2} (x) = 1sec

2

(x)−tan

2

(x)=1

Hope it helps you....!