Question

prove that
sinA×secA+cosA×cosecA=secA×cosecA​

Answer 1

Question :

To prove:

sinA × secA+ cos A × cosecA = secA × cosec A

Trignometric Formulas

1. sin²A + cos²A = 1
2. sec²A - tan²A = 1
3. cosec²A - cot²A = 1
4. cosecx =$\frac{1}{sinx}$
5. secx = $\frac{1}{cosx}$

Solution :

LHS

$= \sin \: A \: \times sec \: A + \cos \: A\times cosec \: A$

$= sin \: A \times \frac{1}{cos \: A} + cos \: A\: \times \frac{1}{sin \: A}$

$= \frac{sin \: A}{cos \: A} + \frac{cos \: A}{sin \: A}$

$= \frac{ \sin {}^{2} \: A+ cos {}^{2} a}{sin \: A\: \times cos \: A}$

we know that sin²A + cos²A = 1

$= \frac{1}{cos \: A\: \times sin \: A}$

$= sec \: A\times cosec \: A$

RHS =secA × cosec A

_____________

⇒LHS = RHS

$\huge{\bold{ Hence\: Proved}}$

__________________________

More Trigonometry Formulas

1. sin2A = 2 sinA cosA
2. cos2A = cos²A - sin²A
3. tan2A = 2 tanA / (1 - tan²A)

Answer 2

Answer:

$\huge{\boxed{\green{\mathcal{Answer-}}}}$

$\sin(a) \times \sec(a) + \cos(a) \times \csc(a) = \sec(a) \times \csc(a)$

♡ FORMULA USED :

${ \sin(a) }^{2} + { \cos(a) }^{2} = 1$

$\sin(a) = \frac{1}{ \csc(a) }$

$\cos(a) = \frac{1}{ \sec(a) }$

TAKING L.H.S.

----》

L.H.S.

$= \sin(a) \times \csc(a) + \cos(a) \times \csc(a) \\ = \frac{ \sin(a) }{ \cos(a) } + \frac{ \cos(a) }{ \sin(a) } \\ = \frac{ { \sin(a) }^{2} + { \cos(a) }^{2} }{ \cos(a) \times \sin(a) } \\ \ = \frac{1}{ \cos(a) \times \sin(a) } \\ = \sec(a) \times \csc(a)$

= R.H.S.

HOPE THAT HELPS