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sergejj [24]
3 years ago
13

Verify identity: (sec(x)-csc(x))/(sec(x)+csc(x))=(tan(x)-1)/(tan(x)+1)

Mathematics
1 answer:
Nikitich [7]3 years ago
7 0
So hmmm let's do the left-hand-side first

\bf \cfrac{sec(x)-csc(x)}{sec(x)+csc(x)}\implies \cfrac{\frac{1}{cos(x)}-\frac{1}{sin(x)}}{\frac{1}{cos(x)}+\frac{1}{sin(x)}}\implies 
\cfrac{\frac{sin(x)-cos(x)}{cos(x)sin(x)}}{\frac{sin(x)+cos(x)}{cos(x)sin(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)sin(x)}\cdot \cfrac{cos(x)sin(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}

now, let's do the right-hand-side then  

\bf \cfrac{tan(x)-1}{tan(x)+1}\implies \cfrac{\frac{sin(x)}{cos(x)}-1}{\frac{sin(x)}{cos(x)}+1}\implies \cfrac{\frac{sin(x)-cos(x)}{cos(x)}}{\frac{sin(x)+cos(x)}{cos(x)}}
\\\\\\
\cfrac{sin(x)-cos(x)}{cos(x)}\cdot \cfrac{cos(x)}{sin(x)+cos(x)}\implies \boxed{\cfrac{sin(x)-cos(x)}{sin(x)+cos(x)}}

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In a right triangle one leg measures a quarter of the other, how much do its acute angles measure? with procedure please​
KATRIN_1 [288]

Answer:

The answer to your question is 14° and 76°

Step-by-step explanation:

Data

leg 1 = x

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angle 1 = ?

angle 2 = ?

Process

To solve problem use trigonometric functions. We must use the trigonometric function that relates both legs (tangent).

    tangent Ф = Opposite side / Adjacent side

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     Ф = tan⁻¹(1/4) =  14°

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