Question

Prove this identity
1/(cosA+sinA) +1/(cosA-sinA)=tan2AcosecA

Answer 1

take LCM and then simplify...

Answer 2

CosA+sinA+CosA-sinA/Cos^2A-CosAsinA+CosASinA-Sin^2A
2CosA/Cos^2A-Sin^2A=2tanA/1-tan^2A

Answer 3

2cosA/Cos2A

Answer 4

plz c my attachment

Answer 5

Thanks @shamim but i didnt understand the part from 2cosA/Cos2A

Answer 6

i took sinA as both numerator n denominator

Answer 7

we can cancell sinA frm denominator n numerator

Answer 8

i think u know
sin2A=2sinA.cosA

Answer 9

u know 1/sinA=cosecA

Answer 10

Yes i Do Know that 1/sinA=cosecA and that sin2A=2SinACosA
so @shamim did you simplify the right side into 2cosA/Cos2A=sin2A/Cos2A*1/Sin

Answer 11

ya i need to get sin2A as numerator

Answer 12

Multiply the left hand side numerator and denominator by SinA
2CosAsinA/Cos2ASinA

Answer 13

ya

Answer 14

\[\frac{1}{(cosA+sinA)} +\frac{1}{(cosA-sinA)}=\tan2AcosecA \]
\[LHS=\frac{1}{(cosA+sinA)} +\frac{1}{(cosA-sinA)}\]
\[LHS=\frac{(cosA-sinA) + (cosA+sinA)}{(cosA+sinA) \times (cosA-sinA)} \]
\[\huge \rightarrow LHS=\frac{cosA-sinA + cosA+sinA}{(\cos^2A-\sin^2A) } \]
\[\huge \rightarrow LHS=\frac{2cosA}{\cos2A } \]
Multiplying Numerator and denominator by sin A, we find:
\[\huge \rightarrow LHS=\frac{2cosA \sin A}{\cos2A \sin A} \]
\[\huge \rightarrow LHS=\frac{2\sin AcosA }{\cos2A \sin A} \]
\[\huge \rightarrow LHS=\frac{\sin2A }{\cos2A \sin A} \]
\[\huge \rightarrow LHS=\frac{\sin2A }{\cos2A } \times \frac{1 }{sinA }\]
\[\huge \rightarrow LHS=\tan 2A \times cosec A\]
\[\huge \rightarrow LHS=\tan 2A cosec A = RHS\]

Hence the identity is Proved.

Answer 15

@Abmon98

Answer 16

How Can i give a medal to each of you thank you for your Help @dpasingh and @shamim :D

Answer 17

wc

Answer 18

@Abmon98 please click on the BEST RESPONSE button besude my name

Answer 19

i have already chose best response i wish i could give a medal to both of you for your help.

Answer 20

chosen*