Question

prove that

$\frac{tanx-sinx}{sin^{2} x} =\frac{tanx}{1+cosx}$
please help exam is tommorow
class 10

Answer 1

Answer:

hey here is your solution

pls mark it as brainliest

Step-by-step explanation:

$so \: here \\ tan \: x - cos \: x/sin {}^{2} x = tan \: x/1 + cos \: x \\ let \: lhs = tan \: x - cos \: x/sin {x }^{2} \\ rhs = tanx/1+cos \: x$

$considering \: lhs \\ = tan \: x - sin \: x/sin {}^{2} x \\ =( sin \: x \div cos \: x) - sin \: x/1 - cos {}^{2} x \\ since \: tan \: x = sin \: x/cos \: x \\ sin {}^{2} x = 1 - cos {}^{2} x \\ \\ ie \: sin \: x - cos \: x.sin \: x/cos \: x/(1 + cos \: x)(1 - cos \: x) \\ = sin \: x(1 - cos \: x)/cos \: x(1 + cos \: x)(1 - cos \: x) \\ = sin \: x/cos \: x \times (1 + cos \: x) \\ \\ = sin \: x/cos \: x \times (1 + cos \: x) \\ = tan \: x/1 + cos \: x \\ since \: \: tan \: x = sin \: x/cos \: x$

$thus \: lhs = rhs \\ hence \: proved$