Maths / Introduction to Trigonometry / Reciprocal Trigonometric Ratios

QUESTION

If  $cos\Theta =\frac {p}{q},$ find the value of $cosec\Theta .$

 OPTIONS A. $\dpi{120} \fn_jvn \large \frac {q}{\sqrt {q^2-p^2}}$ B. $\dpi{120} \fn_jvn \large \frac {\sqrt {q^2-p^2}}{p}$ C. Both a and b D. None of these
Right Option : A

EXPLANATION
Explain TypeExplanation Content
Text

We know  $cos\Theta =\frac {base}{hypotenuse}$

We draw a right triangle, with right angle at C and having base BC = p and hypotenuse BA = q.

By the pythagorean theorem

$AB^2=BC^2+AC^2$

$\Rightarrow$                                           $q^2=p^2+AC^2$

$\Rightarrow$                                           $AC^2=q^2+AC^2$

$\Rightarrow$                                          $AC= \sqrt {q^2-p^2}\: \: \: \: \: \: [\therefore AC>0]$

Thus,                           $cosec\Theta =\frac {hypotenuse}{perpendicular }=\frac {q}{\sqrt {q^2-p^2}}$

View Contents
(Concept based Learning and Testing for [6th - 10th], NTSE, Bank & Govt. Exams)
Self Learning
Testimonials
STUDENT FEEDBACK - Jahanvi C/o ABHYAS Academy
10th
I have a wonderful experience in Abhyas academy. I have improved my learning skills habbit of self study

#### Other Testimonials

Courses We Offer
(Concept based Learning and Testing for [6th - 10th], NTSE, Bank & Govt. Exams)