Question

answer the following question with solution and clear explanation ​

Answer 1

cosA(1-3)/cosA + sin(1+3)/sinA

cancel cos and sin,

we get,

1-3+4= 4-2 = 2

Answer 2

SOLUTION:-

Take L.H.S

$= > \frac{ {cos}^{3}A- cos3A }{cos A} + \frac{ {sin}^{3} A + sin3A}{sin A} \\ \\ = > \frac{ {cos}^{3} A - (4 {cos}^{3}A - 3cosA) }{cosA} + \frac{ {sin}^{3} A + 3sin A - 4 {sin}^{3} A}{sin \: A} \\ \\ = > \frac{3cos \: A - 3 {cos}^{ 3} A}{cos A} + \frac{3sin \: A - 3 {sin}^{3} A }{sin \: A } \\ \\ = > 3 - 3 {cos}^{2} A + 3 - 3 {sin}^{2} A \\ \\ = > 6 - 3( {sin}^{2} A + {cos}^{2} A) \\ \\ = > 6 - 3 \times 1 \\ \\ = > 6 - 3 \\ \\ = > 3 \: \: \: \: \: \: \: \: [R.H.S.]$

Hope it helps ☺️