Question

SecΘ(1-sinΘ)(secΘ+tanΘ)=1

Answer 1

$\huge\underline{\overline{\mid{\bold{\red{ANSWER-}}\mid}}}$

1- sinθ(secθ - tanθ) = 1 - sinθ (secθ) + 1 -sinθ ( tanθ)

=secθ - secθ.sinθ + tanθ - tanθ.sinθ

=secθ- tanθ + tanθ - (sin(sqr)θ / cosθ)

= secθ - (sin(sqr)θ / cosθ)

= ( 1/ cosθ)- (sin(sqr)θ/cosθ)

= ( 1 - sin(sqr)θ/ cosθ) [ LCM cosθ]

= ( cos(sqr)θ / cosθ) [ 1 - sin(sqr)θ = cos(sqr)θ]

= cosθ

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

$\blacksquare \:$FORMULA TO BE IMPLEMENTED

• $\displaystyle \: {sin}^{2} Θ + {cos}^{2} Θ = 1$

• $\displaystyle \: {sec} Θ = \frac{1}{cos Θ}$
• $\displaystyle \: {tan} Θ = \frac{ sin Θ}{ {cos} Θ}$

$\blacksquare \:$CALCULATION

$SecΘ(1-sinΘ)(secΘ+tanΘ)$

$= \displaystyle \: SecΘ(1-sinΘ)( \frac{1}{ \cosΘ} + \frac{sinΘ}{cosΘ} )$

$= \displaystyle \: SecΘ(1-sinΘ)( \frac{1 + sinΘ}{cosΘ})$

$= \displaystyle \: \frac{1}{cosΘ} (1-sinΘ)( \frac{1 + sinΘ}{cosΘ})$

$= \displaystyle \: \frac{1}{ {cos}^{2} Θ} (1-sinΘ)({1 + sinΘ})$

$= \displaystyle \: \frac{1}{ {cos}^{2} Θ} (1- {sin}^{2} Θ)$

$= \displaystyle \: \frac{1}{ {cos}^{2} Θ} \times {cos}^{2} Θ$

$= 1$