Question

show that tan squared theta minus one by cos square theta is equals to minus one​

Answer 1

Step-by-step explanation:

RTP

tan^a - 1/cos^a = -1

Proof

LHS

sin^a/cos^a - 1/cos^ a

= (sin^a-1)/cos^a

= -cos^a/cos^a

= -1

LHS=RHS

Hence Proved

Answer 2

$\tan^2\theta - \frac{1}{\cos^2\theta}$ = -1

Step-by-step explanation:

LHS = $\tan^2\theta - \frac{1}{\cos^2\theta}$

RHS = -1

solving LHS

$\tan^2\theta - \frac{1}{\cos^2\theta}$

we know that

$tan^2\theta= \frac{sin^2\theta }{\cos^2\theta}$

therefore ,

$=\frac{sin^2\theta }{\cos^2\theta}- \frac{1}{\cos^2\theta}\\\\=\frac{sin^2\theta-1}{\cos^2\theta}$

also , $\sin^2\theta+\cos^2\theta=1$

$= \frac{-\cos^2\theta}{\cos^2\theta}\\\\=-1$

= RHS

hence , LHS = RHS

hence proved

#Learn more:

Prove that

cos theta + cos 2theta + cos 3theta + cos 4theta / sin theta + sin 2theta + sin 3theta + sin 4theta = cot 5theta/2​

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