Question

plz do both the questions..


from trigonometry
class 10th​

Answer 1

Step-by-step explanation:

Answer is in the pic.

Hope it will help you!

Answer 2

Step-by-step explanation:

$(i) \: \frac{sin \theta - 2 {sin}^{3} \theta }{2 {cos}^{3} \theta -cos \theta } = tan \theta \\ \\ LHS = \frac{sin \theta - 2 {sin}^{3} \theta }{2 {cos}^{3} \theta -cos \theta } \\ \\ = \frac{sin \theta(1 - 2 {sin}^{2} \theta) }{cos \theta(2 {cos}^{2} \theta -1)} \\ \\ = \frac{sin \theta \: \times cos2 \theta}{cos \theta \: \times cos2 \theta} \\ \\ = \frac{sin \theta }{cos \theta } \\ \\ = tan \theta \\ \\ =RHS \\ \\ thus \: proved \\ \\ (ii) \: {cos}^{2} \theta + {cos}^{2} \theta .{cot}^{2} \theta = {cot}^{2} \theta \\ \\ LHS = {cos}^{2} \theta + {cos}^{2} \theta .{cot}^{2} \theta \\ \\ = {cos}^{2} \theta (1+ {cot}^{2} \theta) \\ \\ = {cos}^{2} \theta {cosec}^{2} \theta \\ \\ = {cos}^{2} \theta \times \frac{1}{{sin}^{2} \theta} \\ \\ = \frac{{cos}^{2}\theta}{{sin}^{2} \theta} \\ \\ = {cot}^{2}\theta \\ \\ =RHS \\ \\ thus \: proved$