Question

If sec X cosec x = 2, then tan^n X + cot^n X equal to ?

Answer 1

Answer:

plzzz give me brainliest ans and plzzzz follow me please

Answer 2

Answer:

${tan}^{n} x + {cot}^{n} x = 2$

Step-by-step explanation:

As we know that

$sec \: x = \frac{1}{cos \: x} \\ \\cosec \: x = \frac{1}{sin \: x} \\ \\ if \: sec \: x \: cosec \: x = 2 \\ \\ \frac{1}{cos \: x} \times \frac{1}{sin\: x} = 2 \\ \\ cross \: multiplying \\ \\ 1 = 2 \: sin \: x \: cos \: x \\ \\ we \: know \: that \\ \\ 2 \: sin \: x \: cos \: x = sin \: 2x \\ \\ sin \: 2x = 1 \\ \\ sin \: 2x = sin \: 90° \\ \\ 2x = 90° \\ \\ x = \frac{90°}{2} \\ \\ x = 45°$

We know that

$tan \: 45° = 1 \\ \\ cot \: 45° = 1 \\ \\ so \\ \\ = {tan}^{n} x + {cot}^{n} x \\ \\ = {(tan \: x)}^{n} + {(tan \: x)}^{n} \\ \\ = {tan}^{n} 45° + {tan}^{n} 45° \\ \\ = {(1)}^{n} + {(1})^{n} \\ \\ = 1 + 1 \\ \\ = 2$

Hope it helps you.