Question

prove (1+cot theta)^2+ (1+tan theta )^2 =(cosec theta + sec theta )
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Answer 1

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$\large\underline{\purple{\sf \orange{\bigstar}Correct\:\: Question:-}}$

$\qquad\qquad\qquad\bf Prove \:\:that :-$

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

$\large\underline{\purple{\sf \orange{\bigstar}Formulae\:and\: Identities\:Used:-}}$

• $\sf (a+b)^2=a^2+b^2+2ab$

• $\sf sin\theta=\dfrac{1}{cosec\theta}$

• $\sf cos\theta=\dfrac{1}{sec\theta}$

• $\sf sin^2\theta+cos^2\theta=1$

• $\sf cosec^2\theta-cot^2\theta=1$

• $\sf sec^2\theta+tan^2\theta=1$

$\large\underline{\purple{\sf\orange{\bigstar} Proof:-}}$

${\red{\bf LHS=}}$

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

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

$=\tt 1+sec^2\theta-1+2tan\theta+cosec^2\theta-1+2cot\theta+2cot\theta$

=$\tt sec^2\theta+cosec^2\theta+2cot\theta+2tan\theta$

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

=$\tt sec^2\theta + cosec^2\theta+2\bigg\lgroup\dfrac{sin\theta}{cos\theta}+\dfrac{cos\theta}{sin\theta}\bigg\rgroup$

=$\tt sec^2\theta + cosec^2\theta+2\bigg\lgroup\dfrac{sin^2\theta+cos^2\theta}{sin\theta.cos\theta}\bigg\rgroup$

=$\tt sec^2\theta + cosec^2\theta+2\bigg\lgroup\dfrac{1}{sin\theta.cos\theta}\bigg\rgroup$

=$\tt sec^2\theta + cosec^2\theta+2(sec\theta.cosec\theta)$

=$\tt sec^2\theta + cosec^2\theta+2(sec\theta.cosec\theta)$

=$\tt (sec\theta+cosec\theta)^2$

${\red{\bf = RHS}}$

$\Large{\boxed{\pink{\sf Hence\: Proved !}}}$

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$\large\underline{\bf\color{blue}{\bigstar}{\blue {More\: Related\: Formulae:-}}}$

➳ $\sf\red{ sin(A+B)=sinA.cosB+cosA.sinB}$

➳ $\sf\red{ sin(A-B)=sinA.cosB-cosA.sinB}$

➳ $\sf\red{cos(A+B)=cosA.cosB-sinA.sinB }$

➳ $\sf\red{cos(A-B)=cosA.cosB+sinA.sinB }$

➳ $\sf\red{ tan(A+B)=\dfrac{tanA+tanB}{1-tanA.tanB}}$

➳ $\sf\red{ tan(A-B)=\dfrac{tanA-tanB}{1+tanA.tanB}}$

➳ $\sf\red{ tan(A+B+C)=\dfrac{tanA+tanB+tanC-tanA.tanB.tanC}{1-tanA.tanB-tanB.tanC-tanC.tanA}}$

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