Question

4) 1 + tan2θ = ?

(A) cot2 θ (B) cosec2 θ (C) sec2 θ (D) tan2 θ​

Answer 1

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$\sf{1~tan²\theta~=~cot²\theta}$

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Answer 2

Answer :

$\longrightarrow \bf \large \red{1 +tan^2\theta =sec^2\theta }$

Solution :

To Find :

• $\sf 1+ tan^2\theta$ .

We know that,

$\red\bigstar \boxed{\sf sin^2\theta+cos^2\theta=1} \sf ....(1)$

Dividing (1) by $cos^2\theta$. we got,

$\sf \dfrac{sin^2\theta}{cos^2\theta}+\cancel{\dfrac{cos^2\theta}{cos^2\theta}}= \dfrac{1}{cos^2\theta}$

We know that ,

$\red\bigstar \boxed{ \sf \dfrac{sin\theta}{cos\theta} = tan\theta} \:\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\ \red\bigstar \boxed{ \sf \dfrac{1}{cos\theta} = sec\theta}$

Appling this here,

$\leadsto \sf tan^2\theta + 1 = sec^2\theta$

Hence,

$\green\star\sf tan^2\theta + 1 = sec^2\theta$

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