Question

prove that 1 + tan theta square + 1 + cot theta square equal to sec theta + cosec theta square ​

Answer 1

Answer:

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Answer 2

$(1+\tan \theta)^2+(1+\cot \theta)^2=(\sec \theta+\csc\theta)^2$, proved.

Step-by-step explanation:

To prove that, $(1+\tan \theta)^2+(1+\cot \theta)^2=(\sec \theta+\csc\theta)^2$.

L.H.S. = $(1+\tan \theta)^2+(1+\cot \theta)^2$

= $1+\tan^2 \theta+2\tan \theta+1+\cot^2 \theta+2\cot \theta$

Using the algebraic identity,

$(a+b)^{2}=a^{2}+b^{2}+2ab$

= $(1+\tan^2 \theta)+2\tan \theta+(1+\cot^2 \theta)+2\cot \theta$

Using the trigonometric identity,

$\sec^2 \theta-\tan^2 \theta=1$

⇒ $\sec^2 \theta=1+\tan^2 \theta$ and

$\csc^2 \theta-\cot^2 \theta=1$

⇒ $\csc^2 \theta=1+\cot^2 \theta$

= $\sec^2 \theta+\csc^2 \theta++2(\tan \theta+\cot \theta)$

= $\sec^2 \theta+\csc^2 \theta+2(\dfrac{\sin \theta}{\cos \theta} +\dfrac{\cos \theta}{\sin \theta})$

= $\sec^2 \theta+\csc^2 \theta+2\dfrac{\sin^2 \theta+\cos^2 \theta}{\sin \theta\cos \theta}$

Using the trigonometric identity,

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

= $\sec^2 \theta+\csc^2 \theta+2\dfrac{1}{\sin \theta\cos \theta}$

= $\sec^2 \theta+\csc^2 \theta+2\sec \theta\csc \theta$

$= (\sec \theta+\csc\theta)^2$

= R.H.S., proved.

Thus, $(1+\tan \theta)^2+(1+\cot \theta)^2=(\sec \theta+\csc\theta)^2$, proved