Question

prove that (sintheta+sectheta)2+(costheta+cosectheta)2=(1+sectheta.cosectheta)2

Answer 1

Solution: (Instead of θ, I use A)

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To prove :

(sin A + sec A)² + (cos A + cosec A)² = (1 + sec A cosec A)

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Proof :

We know that,

cosec A = $\frac{1}{sin A}$

sec A = $\frac{1}{cos A}$

Substituting the value of cosec A & sec A in Identity.,.

⇒ $(sinA + \frac{1}{cos A} )^2 + (cos A + \frac{1}{sin A} )^2 = (1+ \frac{1}{cosA}( \frac{1}{sinA} )$

⇒ $(sinA + \frac{1}{cos A} )^2 + (cos A + \frac{1}{sin A} )^2 = (1+ \frac{1}{cosAsinA})$

Bringing RHS Through LHS to prove they are equal,

LHS = $(sinA + \frac{1}{cos A} )^2 + (cos A + \frac{1}{sin A} )^2$

⇒ $(\frac{sin A cos A+1}{cos A} )^2 + (\frac{sin A cos A + 1}{sinA} )$

⇒ $\frac{(sinAcosA+1)^2}{(cos^2 A)} + \frac{(sinAcosA+1)^2}{(sin^2 A)}$

⇒ $\frac{(sinAcosA+1)^2(sin^2A)+(sinAcosA+1)^2(cos^2A)}{(cos^2A)(sin^2A)}$


⇒ $\frac{(sinAcosA+1)^2(sin^2A+cos^2A)}{(cos^2A)(sin^2A)}$


We know that,


sin²A + cos²A = 1,


Substituting it here,,


We get,

⇒ $\frac{(sinAcosA+1)^2(1)}{(cos^2A)(sin^2A)}$


⇒  $\frac{(sinAcosA+1)^2}{(cos^2A)(sin^2A)}$


⇒ $\frac{(sinAcosA)^2+ 2sinAcosA + 1}{cos^2Asin^2A}$


⇒ $\frac{sinA^2cosA^2+ 2sinAcosA + 1}{cos^2Asin^2A}$


⇒ $1 + \frac{2sinAcosA + 1}{cos^2Asin^2A}$


⇒ $1 + \frac{2sinAcosA}{cos^2Asin^2A} + \frac{1}{cos^2 A sin^2A}$


⇒ $1 + \frac{2}{sinAcosA} + \frac{1}{sin^2 Acos^2A}$


We know that,


a² + 2ab + b² = (a+b)²


hence,

⇒ $1 + 2 (\frac{1}{sinAcosA}) + (\frac{1}{sin AcosA})^2$


⇒ $1 + 2 ( \frac{1}{cosA}( \frac{1}{sinA} ) ) + (\frac{1}{cos A})^2 (\frac{1}{sinA} )^2$


⇒ $1 + 2 (secA)( cosecA ) + (secA})^2 (cosecA )^2$


⇒ (1 + sec A cosec A )² = RHS


                         ∴ Hence, proved,.

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