Question

22. Show that (tan A x sin A) + cos A = secA​

Answer 1

Answer:

L.H.S=(tanA×sinA)+cosA

={(sinA/cosA)×sinA}+cosA

={(sin²A)/cosA}+cosA

= (sin²A+cos²A)/cosA

= 1/cosA

= secA

R.H.S= secA

now, L.H.S= R.H.S

Explanation:

Formula–

1) sin²A+cos²A=1

2) tanA= sinA/cosA

3) 1/cosA= tanA

hope it helps you☺️

Answer 2

Given :

• Show that :

$\bullet \bf \: (tanx \: \times \: sinx) + cosx = secx$

To Prove :

$\bullet \bf \: (tanx \: \times \: sinx) + cosx = secx$

Formula :

$\\ \bullet \sf \: tanx = \frac{sinx}{cosx}$

$\\ \bullet \sf \: cotx = \frac{cosx}{sinx}$

$\bullet \sf \: {tan}^{2} x + {sec}^{2} x = 1$

$\bullet \sf \: {sin}^{2} x + {cos}^{2} x = 1$

$\bullet \sf \: {cosec}^{2} x + {cot}^{2} x = 1$

Solution :

LHS :

$\implies \sf \: (tanx \: \times \: sinx) + cosx \\ \\ \implies \underline{ \red{ \bf{using \: identities}}} \\ \\ \implies \sf \: \bigg( \frac{sinx}{cosx} \times sinx \bigg) + cosx \: \\ \\ \implies \sf \: \bigg( \frac{ {sin}^{2} x}{cosx} \bigg) + cosx \\ \\ \implies \sf \: \bigg( \frac{ {sin}^{2} x + {cos}^{2} x}{cosx} \bigg) \\ \\ \implies \sf \bigg( \frac{1}{cosx} \bigg) \\ \\ \implies \boxed{ \sf{secx}}$

LHS=RHS

$\therefore \sf \: (tan x \: \times sinx) + cosx = secx$