1/secA+tanA -1/cosA = 1/cosA - 1/secA-tanA
Answers
Answered by
3
Answer:
LHS
1/(secA + tanA) - 1/cosA
=1/(1/cosA+sinA/cosA) -1/cosA
=cos^2A-1-sinA/cosA(1+sinA)
=1-sin^2A-1-sinA/cosA(1+sinA)
= -sinA(1+sinA)/cosA(1+sinA)
=-sinA/cosA = -tanA
RHS
1/cosA - 1/secA-tanA
=1/cosA - cosA/1-sinA
=1-sinA-cos^2A/cosA(1-sinA)
=1-sinA-1+sin^2A/cosA(1-sinA)
=-sinA(1-sinA)/cos(1-sinA)
-sinA/cosA =-tanA
Here LHS=RHS.Hence it is proved.
Answered by
3
Answer:
number du kya?????????
Similar questions