Question

Prove that: (sinθ+cosecθ)2+ (cosθ+secθ)2= 7+tan2

θ+cot2

θ .

Answer 1

Hi ,

Here I am using A instead of theta.

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We know the trigonometric

identities ,

1 ) sin² A + cos² A = 1

2 ) sec² A = 1 + tan² A

3 ) cosec² A = 1 + cot² A

And

4 ) sinA cosecA = 1

5 ) cosA secA = 1

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Now ,

LHS = (sinA+cosecA)² + (cosA+secA)²

= sin²A+cosec² A+2sinAcosecA+

cos² A + sec² A + 2cosAsecA

= (Sin²A + cos² A ) + 2 + 2 + cosec² A

+ Sec² A

= 1 + 2 + 2 + ( 1 + cot² A )+( 1 + tan²A )

= 7 + cot² A + tan² A

= RHS

I hope this helps you.

: )

Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

LHS :-

(sinθ + cosecθ)² + (cosθ+ secθ)²

Furthur,

Using Identity

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

= (sin²θ + cosec²θ + 2sinθcosecθ) + (cos²θ + sec²θ + 2cosθsecθ)

As we know that

${\boxed{\bigstar{{sin\theta cos\theta=1\;and\;cos\theta sec\theta=1}}}}$

= (sin²θ + cosec²θ + 2) + (cos²θ + sec²θ + 2)

Also,

= (sin²θ + cos²θ) + 4 + (cosec²θ + sec²θ)

= 1 + 4 + (1 + cot²θ) + (1 + tan²θ)

= (7 + tan²θ + cot²θ)

LHS = RHS

$\boxed{\begin{minipage}{11 cm} Fundamental Trignometric Indentities \\ \\ \tan (90 - A) = cotA \\ \\ cot (90 - A) = tanA \\ \\ sec (90 - A) = cosecA \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \\ \\ cosec (90 - A) = secA \\ \\ sin^{2}\theta+\cos^{2}\theta =1\\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \end{minipage}}$

$\boxed{\begin{minipage}{11 cm} Fundamental Trignometric Indentities \\ \\ \sin^{2}\theta+\cos^{2}\theta =1 \\ \\ 1+tan^{2}\theta=\sec^{2}\theta \\ \\ 1 + cot^{2}\theta=\text{cosec}^2\theta \\ \\ tan\theta =\dfrac{sin\theta}{cos\theta} \\ \\ cot\theta =\dfrac{cos\theta}{sin\theta} \end{minipage}}$