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Answer 1 · 2017-08-06T11:40:16

Refer to the Explanation.

We have,

#x^2-y^2=(asectheta+btantheta)^2-(atantheta+bsectheta)^2,#

#=a^2sec^2theta+2(ab)secthetatantheta+b^2tan^2theta#

#-a^2tan^2theta-2(ab)tanthetasectheta-b^2sec^2theta,#

#=a^2(sec^theta-tan^2theta)-b^2(sec^2theta-tan^2theta),#

#=a^2*1-b^2*1,#

#=a^2-b^2.#

Hence, the Proof.

Answer 2 · 2018-01-20T04:52:49

Please refer to a Proof in the Explanation.

Here is another Method to prove the assertion :

#x=asectheta+btantheta, and y=atantheta+bsectheta#.

#:. x+y=a(sectheta+tantheta)+b(tantheta+sectheta)#,

# rArr x+y=(a+b)(sectheta+tantheta)#.

Similarly, #x-y=(a-b)(sectheta-tantheta)#.

Therefore,

#(x+y)(x-y)=(a+b)(a-b)(sectheta+tantheta)(sectheta-tantheta)#,

#=(a^2-b^2)(sec^2theta-tan^2theta)#,

#=(a^2-b^2)(1)#.

#rArr x^2-y^2=a^2-b^2#.

Hence, the Proof.