Question

sin A + cos A / sin A - cos A + sin A - cos A / sin A + cos A = 2/ sin^2 A - cos^2 = 2/ 1- 2cos^2 A​

Answer 1

Answer:

1. LHS={(sinA+cosA)/(sinA-cosA)}+{(sinA-cosA)/+sinA+cosA)}
2. {(sinA+cosA)(sinA+cosA)+(sinA-cosA)(sinA-cosA)}/{(sin^2A-cos^2A)}
3. hint:-/{(sin^2A-cos^2A)} by using (a+b)(a-b)=(a^2-b^2)
4. {(sinA+cosA)^2+(sinA-cosA)^2 }/{(sin^2A-cos^2A)}
5. {sin^2A+cos^2A+2sinAcosA+sin^2A+cos^2A-2sinAcosA}/{(sin^2A-cos^2A)}
6. 2/{(sin^2A-cos^2A)}
7. from step5 we know sin^2A+cos^2A=1
8. from this2/{(sin^2A-cos^2A)}
9. we can write sin^2A=1-cos^2A
10. 2/{(sin^2A-cos^2A)}=2/{(1-cos^2A-cos^2A)}
11. =2/{(1-2cos^2A)}