Please solve q 20?

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2 Answers
May 4, 2018

I got it to within the sign, #tan theta = {1-x^2}/{|2x|} #, so rather than belabor it, let's call it choice (D).

Explanation:

#x = sec theta + tan theta#

#x = { 1 + sin theta}/ cos theta#

All the answers are of the form # {x^2 \pm 1}/{kx}# so let's square # x#:

# x ^2 = {1 + 2 sin theta + sin ^2 theta}/{cos ^2 theta}#

# x ^2 = {1 + 2 sin theta + sin ^2 theta}/{1 - sin ^2 theta}#

Let #s = sin theta#

#x^2 - x^2 s^2 = 1 + 2s + s^2#

#(1 + x^2)s^2 + 2s + (1-x^2) = 0#

That factors!

# (s + 1) ((1+ x^2 )s + (1- x^2 )) = 0#

# s= -1 or s = {1-x^2}/{1+x^2} #

#sin theta=-1# means #theta=-90^circ# so the cosine is zero and #sec theta + tan theta# is undefined. So we can ignore that and conclude

#sin theta = {1-x^2}/{1+x^2} #

That's a right triangle whose remaining side is

# \sqrt{ (1+x^2)^2 - (1-x^2)^2 } = sqrt{2(2x^2)} = |2x|#

So

#tan theta = pm {1-x^2}/{2x} #

We could worry about the absolute value, but let's just call this choice #D.#

May 4, 2018

Option (D).

Explanation:

Given that, #sectheta+tantheta=x......(1)#.

We know that, #sec^2theta-tan^2theta=1#.

#:. (sectheta+tantheta)(sectheta-tantheta)=1#.

#:. x(sectheta-tantheta)=1#.

#:. sectheta-tantheta=1/x......(2)#.

#:. (1)-(2) rArr 2tantheta=x-1/x=(x^2-1)/x#.

# rArr tantheta=(x^2-1)/(2x)#.

Hence, option (D).