Question

prove that whole under root 1 minus sin theta upon 1 + sin theta = sec theta minus 10 theta

Answer 1

hope this would work......

Answer 2

multiply and divide by 1 + sin theta inside the square root

u will get [1 - sin square theta] / [1 + sin theta]^2 whole under root

which is equal to cos theta  / 1 + sin theta  .....  as 1 - sin square = cos square theta

take its reciprocal and split the terms

u will get [1/ cos theta] + [sin theta / cos theta] which is equal to sec theta + tan theta

now [sec theta + tan theta][sec theta -tan theta] = 1

therefore sec theta + tan theta = 1/[sec theta - tan theta]
 
now take its reciprocal

hence proved