Question

Prove that
$\frac{ \sin + \cos - 1 }{ \sin + \cos + 1 } = \frac{1}{ \sec + \tan }$
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Answer 1

Answer:

Recall the definitions of the reciprocal trigonometric functions, csc θ, sec θ and cot θ from the trigonometric functions chapter:

\displaystyle \csc{\theta}=\frac{1}{{ \sin{\theta}}}cscθ=

sinθ

1

\displaystyle \sec{\theta}=\frac{1}{{ \cos{\theta}}}secθ=

cosθ

1

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

tanθ

1

Now, consider the following diagram where the point (x, y) defines an angle θ at the origin, and the distance from the origin to the point is r units:

Angle on cartesian plane

From the diagram, we can see that the ratios sin θ and cos θ are defined as:

\displaystyle \sin{\theta}=\frac{y}{{r}}sinθ=

r

y

and

\displaystyle \cos{\theta}=\frac{x}{{r}}cosθ=

r

x

Now, we use these results to find an important definition for tan θ:

\displaystyle\frac{{ \sin{\theta}}}{{ \cos{\theta}}}=\frac{{\frac{y}{{r}}}}{{\frac{x}{{r}}}}=\frac{y}{{r}}\times\frac{r}{{x}}=\frac{y}{{x}}

cosθ

sinθ

=

r

x

r

y

=

r

y

×

x

r

=

x

y

Now also \displaystyle \tan{\theta}=\frac{y}{{x}}tanθ=

x

y

, so we can conclude:

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

cosθ

sinθ

Ratios based on Pythagoras' Theorem

Also, for the values in the above diagram, we can use Pythagoras' Theorem and obtain:

y2 + x2 = r2

Dividing through by r2 gives us:

\displaystyle\frac{{y}^{2}}{{r}^{2}}+\frac{{x}^{2}}{{r}^{2}}={1}

r

2

y

2

+

r

2

x

2

=1

so we obtain the important result:

\displaystyle{{\sin}^{2}\ }\theta+{{\cos}^{2}\ }\theta={1}sin

2

θ+cos

2

θ=1

Step-by-step explanation:

hope it helps you.