Question

(sino+coseco)^2 + (coso+seco)^2 = 7 + tan^2o +cot^2o​

Answer 1

To prove :

$\sf{(sin\theta+cosec\theta)^2+(cos\theta+sec\theta)^2=7+tan^2\theta+cot^2\theta}$

Proof :

Taking LHS

$:\implies\sf{(sin\theta+cosec\theta)^2+(cos\theta+sec\theta)^2}\\\\ \small{\sf{using\:algebraic\:identity }} \\ \small{\tt{(a+b)^2=a^2+b^2+2ab}}\\\\:\implies\sf{sin^2\theta+cosec^2\theta+2sin\theta cosec\theta +}\\\sf{\:\;\:\:\;\;\;\;\;\;\;cos^2\theta+sec^2\theta+2cos\theta sec\theta}\\\\\small{\sf{using \:trigonometric \:ratios}}\\\small{\sf{sinA\;cosecA=1\;\;and\;\; cosA\;secA=1}}\\\\:\implies\sf{(sin^2\theta+cos^2\theta)+sec^2\theta+cosec^2\theta+2(1)+2(1)}$

$\small{\sf{using\:trigonometric\:identities}}\\\small{\sf{sin^2A+cos^2A=1\:\:, 1+tan^2A=sec^2A}}\\\small{\sf{and\:1+cot^2A=cosec^2A}}\\\\:\implies\sf{(1)+1+tan^2\theta+1+cot^2\theta+2+2}\\\\:\implies\sf{7+tan^2\theta+cot^2\theta}$

= RHS

Proved.

$\rule{205}3$

other trigonometric ratios and values

$\star\;\;\sf{cosecA=\frac{1}{sinA}}$

$\star\;\;\sf{secA=\frac{1}{cosA}}$

$\star\;\;\sf{tanA=\frac{sinA}{cosA}}$

$\star\;\;\sf{cotA=\frac{cosA}{sinA}=\frac{1}{tanA}}$

$\rule{205}1$

$\star\;\;\sf{sin(90-A)=cosA}$

$\star\;\;\sf{cos(90-A)=sinA}$

$\star\;\;\sf{sec(90-A)=cosecA}$

$\star\;\;\sf{cosec(90-A)=secA}$

$\star\;\;\sf{tan(90-A)=cotA}$

$\star\;\;\sf{cot(90-A)=tanA}$

$\rule{205}3$