Question

I hve tried many times bt answer is not coming plz whoever will tell the answer will be marked as BRAINLIEST
I am giving guarantee plz prove that q​

Answer 1

Step-by-step explanation:

${sec}^{4} \theta(1 - {sin}^{4} \theta) - 1 = \frac{2}{{cot}^{2} \theta} \\ LHS = {sec}^{4} \theta(1 - {sin}^{4} \theta) - 1 \\ = {sec}^{4} \theta(1 - {sin}^{2} \theta) (1 + {sin}^{2} \theta)- 1 \\ = {sec}^{4} \theta \: {cos}^{2} \theta (1 + {sin}^{2} \theta)- 1 \\ = {sec}^{2} \theta \: (1 + {sin}^{2} \theta)- 1 \\ = {sec}^{2} \theta \: + {sec}^{2} \theta {sin}^{2} \theta- 1 \\ = {sec}^{2} \theta - 1 + \frac{1}{{cos}^{2} \theta } \times {sin}^{2} \theta \\ = {tan}^{2} \theta + \frac{{sin}^{2} \theta}{{cos}^{2} \theta} \\ = {tan}^{2} \theta + {tan}^{2} \theta \\ = 2 {tan}^{2} \theta \\ = 2 \times \frac{1}{ {cot}^{2} \theta} \\ = \frac{2}{ {cot}^{2} \theta} \\ = RHS$

Thus proved...