Question

if sin theta =-7/15 and theta is in the third quadrant then $\sf\dfrac{7cot\theta-24tan\theta}{7cot\theta +24tan\theta} =$​

Answer 1

Correct question :-

if sinθ  = -7/15 and theta is in the third quadrant then find the value of

$\sf{ \dfrac{7cot\theta-24tan\theta}{7cot\theta+24tan\theta} }$

$\underline{\underline{\sf{\color{magenta}{GIVEN:-}}}}$

$\bullet\ \sf{sin\theta=-\dfrac{7}{15} }$

$\bullet$ θ is in the third quadrant.

$\underline{\underline{\sf{\color{magenta}{TO\ FIND:-}}}}$

The value of

$\sf{\longrightarrow \dfrac{7cot\theta-24tan\theta}{7cot\theta+24tan\theta} }$

$\underline{\underline{\sf{\color{magenta}{SOLUTION:-}}}}$

Using sin²θ + cos²θ = 1,

$\displaystyle \sf{\bigg(\frac{-7}{25}\bigg)^2 +cos^2\theta=1}$

$\implies\displaystyle \sf{cos^2\theta=1-\frac{49}{625} }$

$\implies\displaystyle \sf{cos^2\theta=\frac{625}{625}-\frac{49}{625} }$

$\displaystyle \sf{\implies cos^2\theta=\frac{576}{625} }$

$\displaystyle \sf{\implies cos\theta=\pm\frac{24}{25} }$

As θ is in third quadrant, so cosθ will be -24/25.

Cos θ is negative in third quadrant,

______________________

$\displaystyle \bf{ tan\theta=\frac{sin\theta}{cos\theta} }$

$\displaystyle \sf{\implies tan\theta=\frac{-7}{25}\div \frac{-24}{25} }$

$\displaystyle \sf{\implies tan\theta=\frac{-7}{25}\times\frac{25}{-24} }$

$\displaystyle \sf{\implies tan\theta=\frac{7}{24} }$

_______________________

$\displaystyle \bf{ cot\theta=\frac{1}{tan\theta}}$

$\displaystyle \sf{\implies cot\theta=\frac{24}{7} }$

_______________________

$\displaystyle \sf{\frac{7cot\theta-24tan\theta}{7cot\theta+24tan\theta} }$

$\displaystyle \sf{= \frac{7\times\frac{24}{7}-24\times\frac{7}{24} }{7\times\frac{24}{7}+24\times\frac{7}{24} } }$

$\displaystyle \sf{= \frac{24-7}{24+7} }$

$\displaystyle \sf{= \frac{17}{31} }$

Therefore, the answer is 17/31.

Answer 2

Step-by-step explanation:

Correct question :-

if sinθ = -7/15 and theta is in the third quadrant then find the value of

\sf{ \dfrac{7cot\theta-24tan\theta}{7cot\theta+24tan\theta} }

7cotθ+24tanθ

7cotθ−24tanθ

\underline{\underline{\sf{\color{magenta}{GIVEN:-}}}}

GIVEN:−

\bullet\ \sf{sin\theta=-\dfrac{7}{15} }∙ sinθ=−

15

7

\bullet∙ θ is in the third quadrant.

\underline{\underline{\sf{\color{magenta}{TO\ FIND:-}}}}

TO FIND:−

The value of

\sf{\longrightarrow \dfrac{7cot\theta-24tan\theta}{7cot\theta+24tan\theta} }⟶

7cotθ+24tanθ

7cotθ−24tanθ

\underline{\underline{\sf{\color{magenta}{SOLUTION:-}}}}

SOLUTION:−

Using sin²θ + cos²θ = 1,

\displaystyle \sf{\bigg(\frac{-7}{25}\bigg)^2 +cos^2\theta=1}(

25

−7

)

2

+cos

2

θ=1

\implies\displaystyle \sf{cos^2\theta=1-\frac{49}{625} }⟹cos

2

θ=1−

625

49

\implies\displaystyle \sf{cos^2\theta=\frac{625}{625}-\frac{49}{625} }⟹cos

2

θ=

625

625

−

625

49

\displaystyle \sf{\implies cos^2\theta=\frac{576}{625} }⟹cos

2

θ=

625

576

\displaystyle \sf{\implies cos\theta=\pm\frac{24}{25} }⟹cosθ=±

25

24

As θ is in third quadrant, so cosθ will be -24/25.

Cos θ is negative in third quadrant,

______________________

\displaystyle \bf{ tan\theta=\frac{sin\theta}{cos\theta} }tanθ=

cosθ

sinθ

\displaystyle \sf{\implies tan\theta=\frac{-7}{25}\div \frac{-24}{25} }⟹tanθ=

25

−7

÷

25

−24

\displaystyle \sf{\implies tan\theta=\frac{-7}{25}\times\frac{25}{-24} }⟹tanθ=

25

−7

×

−24

25

\displaystyle \sf{\implies tan\theta=\frac{7}{24} }⟹tanθ=

24

7

_______________________

\displaystyle \bf{ cot\theta=\frac{1}{tan\theta}}cotθ=

tanθ

1

\displaystyle \sf{\implies cot\theta=\frac{24}{7} }⟹cotθ=

7

24

_______________________

\displaystyle \sf{\frac{7cot\theta-24tan\theta}{7cot\theta+24tan\theta} }

7cotθ+24tanθ

7cotθ−24tanθ

\displaystyle \sf{= \frac{7\times\frac{24}{7}-24\times\frac{7}{24} }{7\times\frac{24}{7}+24\times\frac{7}{24} } }=

7×

7

24

+24×

24

7

7×

7

24

−24×

24

7

\displaystyle \sf{= \frac{24-7}{24+7} }=

24+7

24−7

\displaystyle \sf{= \frac{17}{31} }=

31

17

Therefore, the answer is 17/31.