Question

theta is an acute angle such that cot square theta =7/8 find the value of (1+sin theta) (1-sin theta)/(1+cos theta )(1-cos theta)​

Answer 1

Answer:

Given that θ is an acute angle which gives us θ lies in first quadrant i.e.

θ € 1st quadrant.

and we know that in first quadrant

sinθ = +ve...........(¡)

&

cosθ =- ve...........(¡¡)

[bold equations needed only if single power of trigonometric ratios is asked].

Given that cot^2θ =7/8

Now as we know that

$1 + {cot}^{2} \theta = {cosec}^{2} \theta \\ \implies \: cosec {}^{2} \theta = 1 + \frac{7}{8} \\ = \frac{15}{8} .$

Also we know that

${cosec}^{2} \theta = \frac{1}{ {sin}^{2}\theta} \\ \implies {sin}^{2} \theta = \frac{8}{15}$

Also,

${cos}^{2} \theta = 1 - {sin}^{2} \theta \\ \implies {cos}^{2} \theta = 1 - \frac{8}{15} \\ \implies {cos}^{2} \theta = \frac{7}{15}$

Now using these values of Cos^2θ and sin^2θ we can easily calculate the value of given expression as:-

$\frac{(1 + sin \theta)(1 - sin \theta) }{(1 + cos \theta)(1 - cos \theta)} \\ = \frac{ {1}^{2} - {sin}^{2} \theta}{ {1}^{2} - {cos}^{2} \theta}$

$= \frac{1 - \frac{8}{15} }{1 - \frac{7}{15} } \\ = \frac{ \frac{7}{15} }{ \frac{8}{15} } \\ = \frac{7}{8} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{ANSWER}} \:$

Answer 2

Answer:

Given that θ is an acute angle which gives us θ lies in first quadrant i.e.

θ € 1st quadrant.

and we know that in first quadrant

sinθ = +ve...........(¡)

&

cosθ =- ve...........(¡¡)

[bold equations needed only if single power of trigonometric ratios is asked].

Given that cot^2θ =7/8

Now as we know that

\begin{gathered} 1 + {cot}^{2} \theta = {cosec}^{2} \theta \\ \implies \: cosec {}^{2} \theta = 1 + \frac{7}{8} \\ = \frac{15}{8} .\end{gathered}

1+cot

2

θ=cosec

2

θ

⟹cosec

2

θ=1+

8

7

=

8

15

.

Also we know that

\begin{gathered} {cosec}^{2} \theta = \frac{1}{ {sin}^{2}\theta} \\ \implies {sin}^{2} \theta = \frac{8}{15} \end{gathered}

cosec

2

θ=

sin

2

θ

1

⟹sin

2

θ=

15

8

Also,

\begin{gathered} {cos}^{2} \theta = 1 - {sin}^{2} \theta \\ \implies {cos}^{2} \theta = 1 - \frac{8}{15} \\ \implies {cos}^{2} \theta = \frac{7}{15} \end{gathered}

cos

2

θ=1−sin

2

θ

⟹cos

2

θ=1−

15

8

⟹cos

2

θ=

15

7

Now using these values of Cos^2θ and sin^2θ we can easily calculate the value of given expression as:-

\begin{gathered} \frac{(1 + sin \theta)(1 - sin \theta) }{(1 + cos \theta)(1 - cos \theta)} \\ = \frac{ {1}^{2} - {sin}^{2} \theta}{ {1}^{2} - {cos}^{2} \theta} \\ \end{gathered}

(1+cosθ)(1−cosθ)

(1+sinθ)(1−sinθ)

=

1

2

−cos

2

θ

1

2

−sin

2

θ

\begin{gathered} = \frac{1 - \frac{8}{15} }{1 - \frac{7}{15} } \\ = \frac{ \frac{7}{15} }{ \frac{8}{15} } \\ = \frac{7}{8} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \boxed{ANSWER}} \: \end{gathered}

=

1−

15

7

1−

15

8

=

15

8

15

7

=

8

7

ANSWER