Question

tanx /secx+1 + secx+1/tanx=
1) 2cosecx 2) 2secx
3) 2sinx. 4)2cosx

Answer 1

Solution :

Now, $\frac{tanx}{secx + 1} + \frac{secx + 1}{tanx}$

= $\frac{tanx*tanx+(secx+1)(secx+1)}{(secx+1)tanx}$

= $\frac{tan^{2}x+sec^{2}x+2secx+1}{(secx+1)tanx}$

= $\frac{(tan^{2}x+1)+sec^{2}x+2secx}{(secx+1)tanx}$

= $\frac{sec^{2}x+sec^{2}+2secx}{(secx+1)tanx}$

{ since $sec^{2}x-tan^{2}x=1$ }

= $\frac{2sec^{2}x+2secx}{(secx+1)tanx}$

= $\frac{2secx(secx+1)}{(secx+1)tanx}$

= $\frac{2secx}{tanx}$

{by cancelling (secx+1) from both the numerator and the denominator}

= $\dfrac{(\frac{2}{cosx})}{(\frac{sinx}{cosx})}$

= $\frac{2cosx}{sinx*cosx}$

= $\frac{2}{sinx}$

= 2 cosecx

Thus, option ( 1 ) is correct

Answer 2

Answer:

+

tanx

secx+1

= \frac{tanx*tanx+(secx+1)(secx+1)}{(secx+1)tanx}

(secx+1)tanx

tanx∗tanx+(secx+1)(secx+1)

= \frac{tan^{2}x+sec^{2}x+2secx+1}{(secx+1)tanx}

(secx+1)tanx

tan

2

x+sec

2

x+2secx+1

= \frac{(tan^{2}x+1)+sec^{2}x+2secx}{(secx+1)tanx}

(secx+1)tanx

(tan

2

x+1)+sec

2

x+2secx

= \frac{sec^{2}x+sec^{2}+2secx}{(secx+1)tanx}

(secx+1)tanx

sec

2

x+sec

2

+2secx

{ since sec^{2}x-tan^{2}x=1sec

2

x−tan

2

x=1 }

= \frac{2sec^{2}x+2secx}{(secx+1)tanx}

(secx+1)tanx

2sec

2

x+2secx

= \frac{2secx(secx+1)}{(secx+1)tanx}

(secx+1)tanx

2secx(secx+1)

= \frac{2secx}{tanx}

tanx

2secx

{by cancelling (secx+1) from both the numerator and the denominator}

= \dfrac{(\frac{2}{cosx})}{(\frac{sinx}{cosx})}

(

cosx

sinx

)

(

cosx

2

)

= \frac{2cosx}{sinx*cosx}

sinx∗cosx

2cosx

= \frac{2}{sinx}

sinx

2

= 2 cosecx

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thank you