Question

Prove that:-
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Answer 1

Answer:

use identity

Step-by-step explanation:

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Answer 2

Question :

Prove that :

$- 1 + \frac{ \sin(A) \ \sin(90 - A) }{ \cot(90 - A) } = - \sin {}^{2} (A)$

Formulas :

• sin (90° - A ) = cosA
• cot (90° -A ) = tan A
• tan ( 90° - A ) = cot A
• sec( 90° - A ) = cosec A
• cosec ( 90° - A )= cot A
• cot ( 90° - A ) = tanx
• $sin {}^ {2} A + cos {}^{2} A = 1$

${\green {\huge{\underline{\mathbb{Answer}}}}}$

LHS = $- 1 + \frac{ \sin(A) \ \sin(90 - A) }{ \cot(90 - A) }$

$= - 1 + \frac{ \sin(A) \cos(A) }{ \tan(A) }$

$= - 1 + \frac{ \sin(A) \cos(A) }{ \frac{ \sin(A) }{ \cos(A) } }$

$= - 1 + \frac{ \sin(A) \times \cos(A) \times \cos(A) }{ \sin(A) }$

$= - 1 + \cos {}^{2} (A)$

$= - ( - \cos {}^{2} (A) + 1)$

$= - \sin {}^{2} (A)$

$= RHS$

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