Question

Decide whether the equations is a trigonometric function identity, EXPLAIN.

cos^2x(1+tan^2x)=1

secxtanx(1-sin^2)=sinx

sin^2(theta)csc^2(theta)=sin^2(theta)+cos^2(theta)

Answer 1

(1) Answer: Identity true

$\cos^2x(1+\tan^2x)=\cos^2x(1+\frac{sin^2x}{cos^2x})=\\= \cos^2x\frac{cos^2x+sin^2x}{cos^2x}=1$

(2) Answer: Identity true

$\sec x \tan x(1-\sin^2x)=\sin x\\\frac{1}{\cos x}\frac{\sin x}{\cos x}(\cos^2x)=\sin x\\\sin x = \sin x$

(3) Answer: Identity true

$\sin^2(\theta)\csc^2(\theta)=\sin^2(\theta)+\cos^2(\theta)\\\sin^2(\theta)\frac{1}{\sin^2(\theta)}=\sin^2(\theta)+\cos^2(\theta)\\1 = 1$

Answer 2

The equations is a trigonometric function identity is true.

Explain whether the equations is a trigonometric function identity or not.

What is trigonometry?

The branch of mathematics concerned with specific functions of angles and their application to calculations.

(i) $cos^{2} x(1+tan^{2} )=cos^{2}x$

$cos^{2} x\frac{cos^{2}x+sin^{2}x }{cos^{2}x } =1$

1=1

LHS=RHS

(ii) $secxtanx(1-sin^{2}x =sinx$

$\frac{1*sinx}{cosx*cosx} (cos^{2}x) =sinx$

$sinx=sinx\\1=1$

LHS=RHS

(iii)

$sin^{2}xcosec^{2}x =sin^{2}x+cos^{2}x\\ sin^{2}x*\frac{1}{sin^{2}x } =sin^{2}x+cos^{2}x\\1=1$

LHS=RHS

So, equations is a trigonometric function identity is true.

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