Question

5 cosec 0 = 7. then evaluate sino + cos 0 - 1​

Answer 1

Answer:

$\bf \: cosec \: \theta \: = \frac{7}{5} = \frac{hypotenuse }{opposite}$

Need to find Adjecent side!

By Pythagoras theorem:

$\bf \: adjecent \: \: side \: \: = \sqrt{ {(hypotenuse \: side)}^{2} - {(opposite \: \: side)}^{2} } \\ \\ \bf \: adjecent \: \: side \: \: = \sqrt{ {(7)}^{2} - {(5)}^{2} } \\ \\ \bf \: adjecent \: \: side \: \: = \sqrt{49 - 25} \\ \\ \bf \: adjecent \: \: side \: \: = \sqrt{24} \\ \\ \bf \: adjecent \: \: side \: \: = \sqrt{6 \times 4} \\ \\ \bf \: adjecent \: \: side \: \: =2 \sqrt{6}$

We know that,

$\bf \sin \theta = \frac{opposite}{hypotenuse \: } = \frac{5}{7} \\ \\ \bf \: cos \: \theta = \frac{adjecent \: }{hypotenuse} = \frac{2 \sqrt{6} }{7}$

now,

$\bf \implies \: sin \: \theta \: + cos \: \theta - 1 \\ \\ \bf \implies \: \frac{5}{7} + \frac{2 \sqrt{6} }{7} - 1 \\ \\ \bf \implies \: \frac{7 \sqrt{6} }{7} - 1 \\ \\ \bf \implies \: \frac{ \cancel7 \: \sqrt{6} }{ \cancel7} - 1 \\ \\ \bf \implies \: \sqrt{6} - 1$

Answer 2

Step-by-step explanation:

Answer:

\bf \: cosec \: \theta \: = \frac{7}{5} = \frac{hypotenuse }{opposite}cosecθ=

5

7

=

opposite

hypotenuse

Need to find Adjecent side!

By Pythagoras theorem:

\begin{gathered}\bf \: adjecent \: \: side \: \: = \sqrt{ {(hypotenuse \: side)}^{2} - {(opposite \: \: side)}^{2} } \\ \\ \bf \: adjecent \: \: side \: \: = \sqrt{ {(7)}^{2} - {(5)}^{2} } \\ \\ \bf \: adjecent \: \: side \: \: = \sqrt{49 - 25} \\ \\ \bf \: adjecent \: \: side \: \: = \sqrt{24} \\ \\ \bf \: adjecent \: \: side \: \: = \sqrt{6 \times 4} \\ \\ \bf \: adjecent \: \: side \: \: =2 \sqrt{6}\end{gathered}

adjecentside=

(hypotenuseside)

2

−(oppositeside)

2

adjecentside=

(7)

2

−(5)

2

adjecentside=

49−25

adjecentside=

24

adjecentside=

6×4

adjecentside=2

6

We know that,

\begin{gathered}\bf \sin \theta = \frac{opposite}{hypotenuse \: } = \frac{5}{7} \\ \\ \bf \: cos \: \theta = \frac{adjecent \: }{hypotenuse} = \frac{2 \sqrt{6} }{7}\end{gathered}

sinθ=

hypotenuse

opposite

=

7

5

cosθ=

hypotenuse

adjecent

=

7

2

6

now,

\begin{gathered}\bf \implies \: sin \: \theta \: + cos \: \theta - 1 \\ \\ \bf \implies \: \frac{5}{7} + \frac{2 \sqrt{6} }{7} - 1 \\ \\ \bf \implies \: \frac{7 \sqrt{6} }{7} - 1 \\ \\ \bf \implies \: \frac{ \cancel7 \: \sqrt{6} }{ \cancel7} - 1 \\ \\ \bf \implies \: \sqrt{6} - 1\end{gathered}

⟹sinθ+cosθ−1

⟹

7

5

+

7

2

6

−1

⟹

7

7

6

−1

⟹

7

7

6

−1

⟹

6

−1