Question

Tantheta+sintheta/tantheta-sintheta=sectheta+1/sectheta-1

Answer 1

Given Equation : $\dfrac{tanA + sinA}{tanA - sinA}$

     Method 1 for the solution of the given equation is given below

⇒ $\dfrac{tanA + sinA}{tanA-sinA}$

From trigonometric identities : tanA = $\dfrac{sinA}{cosA}$

∴ $\dfrac{\dfrac{sinA}{cosA} + sinA}{\dfrac{sinA}{cosA} - sinA}$

Divide numerator & denominator both by sinA

⇒ $\dfrac{\dfrac{\dfrac{sinA}{cosA}}{ \dfrac{sinA}{1}}+ \dfrac{sinA}{sinA}}{\dfrac{\dfrac{sinA}{cosA}}{\dfrac{sinA}{1}} - \dfrac{sinA}{sinA}}$

⇒ $\dfrac{\dfrac{1}{cosA} + 1}{\dfrac{1}{cosA} - 1}$

From trigonometric identities : $\dfrac{1}{cosA} = secA$

⇒ $\dfrac{secA + 1}{secA -1 }$

Hence, proved that $\dfrac{tanA + sinA}{tanA-sinA} = \dfrac{secA+1}{secA-1}$

          Method 2 for the solution of the given equation is given below

⇒ $\dfrac{tanA + sinA}{tanA - sinA}$

From trigonometric identities : tanA = $\dfrac{sinA}{cosA}$

⇒ $\dfrac{\dfrac{sinA}{cosA} + sinA }{ \dfrac{sinA}{cosA} - sinA}$

⇒ $\dfrac{sinA \bigg( \dfrac{1}{cosA} + 1 \bigg) }{sinA \bigg(\dfrac{1}{cosA} - 1\bigg) }$

From trigonometric identities : $\dfrac{1}{cosA} = secA$

⇒ $\dfrac{secA + 1}{secA - 1 }$

Hence, proved that $\dfrac{tanA + sinA}{tanA-sinA} = \dfrac{secA+1}{secA-1}$

Answer 2

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student-name Roshan asked in Math
tan theta + sin theta / tan theta - sin theta = sec theta + 1 / sec theta - 1

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student-name Anish answered this
30 helpful votes in Math, Class X
take # as tita ( i dont know how to input tita on the computer)

Tan#+Sin# / Tan# - Sin# = Sec#+1 / Sec# - 1

L.H.S= Tan#+ Sin# / Tan# - Sin#

= ( Sin# / Cos# +Sin# ) / ( Sin# / Cos# - Sin#)

= (Sin# + Sin#Cos# / Cos#) / ( Sin# - Sin#Cos# / Cos#)

= (Sin# + Sin#Cos#) / (Sin# - Sin#Cos#)

Taking Sin# common on both denominator and numerator (they get cancelled)

We obtain :

1+ Cos# / 1 - Cos#

R.H.S = Sec# + 1 / Sec# - 1

= (1/Cos# +1) / (1/Cos# -1 )

= 1+Cos# / 1 -Cos#

L.H.S = R.H.S

(i really am not good in inputting math answers onto the computer so i apologize to you if you had any hardship in reading the answer)

you don't understand you see that.... OK...