Question

please give the solution for all the questions​

Answer 1

Answer:

10) tan 3A = cot( A-26)

cot (90-3A) = cot (A-26)

90- 3A= A-26

4A= 116

A= 116/4=29

Step-by-step explanation:

12)

$(1 + { \cot}^{2} \alpha ) {sin}^{2} \alpha$

${cosec}^{2} \alpha \times {sin}^{2} \alpha = 1$

13)

${sin}^{2} \alpha + \frac{1}{1 + {tan}^{2} \alpha } \\ { {sin}^{2} \alpha + \frac{1}{ {sec}^{2} \alpha } }^{2} \\ {sin}^{2} \alpha + {cos}^{2} \alpha = 1$

14)sinAcos(90-A)+cosAsin(90-A)

sin A.sinA + cosA.cosA

${sin}^{2}a + {cos}^{2} a = 1$