Question

Trigeonometric Identities

Exercise 13 A

Question 30

(cos ∅ × cosec theta - sin ∅ × sec theta) / (cos ∅ + sin ∅) = cosec ∅ - sec ∅

Those who knows only they will answer​

Answer 1

Answer:

proof

Step-by-step explanation:

cos.cosec-sin.sec/cos+sin

=cos.1/sin-sin.1/cos÷cos+sin

=cos^2-sin^2/sin.cos÷cos+sin

=cos^2-sin^2÷sin.cosxcos+sin

=cos-sin÷sinxcos

=cos÷sin.cos-sin÷sin.cos

=1/cos-1/sin

=cosec-sec

Answer 2

Correct Question :

$\dagger \: \: \: \: \: \tt \dfrac{Cos\phi. \, Cosec\phi - Sin\phi .\, Sec\phi}{Cos\phi + Sin\phi} = Cosec\phi - Sec\phi$

Taking LHS side.

$\tt \longrightarrow \dfrac{Cos\phi. \, Cosec\phi - Sin\phi .\, Sec\phi}{Cos\phi + Sin\phi}$

$\tt \longrightarrow \dfrac{Cos\phi. \, \frac{1}{Sin\phi}- Sin\phi .\, \frac{1}{Cos\phi}}{Cos\phi + Sin\phi}$

$\tt \longrightarrow \dfrac{\, \frac{Cos\phi}{Sin\phi}- \frac{Sin\phi}{Cos\phi}}{Cos\phi + Sin\phi}$

$\tt \longrightarrow \dfrac{\, \frac{Cos^2\phi - Sin^2\phi}{Sin\phi.Cos\phi}}{Cos\phi + Sin\phi}$

$\tt \longrightarrow \dfrac{Cos^2\phi - Sin^2\phi}{(Sin\phi.Cos\phi) Cos\phi + Sin\phi }$

$\tt \longrightarrow \dfrac{Cos\phi - Sin\phi) \cancel{( Cos\phi + Sin\phi)} }{(Sin\phi.Cos\phi) \cancel{ (Cos\phi + Sin\phi )}}$

$\tt \longrightarrow \dfrac{Cos\phi - Sin\phi}{Sin\phi.Cos\phi }$

$\tt \longrightarrow \dfrac{ \: \: \: \: \: \: \: \: \: \cancel{Cos\phi}^{ \: \: \: \: \: \: 1}}{Sin\phi. \cancel{Cos\phi }} - \dfrac{ \: \: \: \: \: \: \: \: \: \: \cancel{Sin\phi}^{ \: \: \: \: \: 1}}{ \cancel{Sin\phi}.Cos\phi }$

$\tt \longrightarrow \dfrac{{1}}{Sin\phi} - \dfrac{ {1}}{ Cos\phi }$

$\tt \longrightarrow Cosec\phi - Sec\phi.$

Hence Proved.