Problem:
Let x be a real number such that secx−tanx=2. Then secx+tanx=
Answer Choices:
A. 0.1
B. 0.2
C. 0.3
D. 0.4
E. 0.5
Solution:
From the identity 1+tan2x=sec2x it follows that 1=sec2x−tan2x= (secx−tanx)(secx+tanx)=2(secx+tanx), so secx+tanx=0.5.
OR
The given relation can be written as cosx1−sinx=2. Squaring both sides yields 1−sin2x(1−sinx)2=4, hence 1+sinx1−sinx=4. It follows that sinx=−53 and that
cosx=21−sinx=21−(−3/5)=54.
Thus secx+tanx=45−43=0.5.