Question

Prove that: (sinθ+cosecθ)2+ (cosθ+secθ)2= 7+tan2θ+cot2θ .

Answer 1

◕Correct Question◕$\:$

Prove that: (sinθ+cosecθ)2+ (cosθ+secθ)2= 7+tan2θ+cot2θ .

◕Step by step explanation◕

⚽ Approach ⚽

>we will start with LHS and make it equal to RHS

Let's begin:-

⚽ATQ⚽

LHS = (sinθ+cosecθ)² + (cosθ+secθ)²

=> sin²θ+cosec² θ+2sinθcosecθ+

cos² θ + sec² θ + 2cosθsecθ

=> (Sin²θ + cos² θ ) + 2 + 2 + cosec² θ

+ Sec² θ

=> 1 + 2 + 2 + ( 1 + cot² θ )+( 1 + tan²θ )

=> 7 + cot² θ + tan² θ

= RHS

Hence,

⚽ L.H.S=R.H.S ⚽ Verified

☬ some of the trigonometric identities used:

>sin² θ + cos² θ = 1

>sec² θ = 1 + tan² θ

>cosec² θ = 1 + cot² θ

>sinθ cosecθ = 1

>cosθ secθ = 1

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✧ I HOPE IT'S HELPFUL ✧

Answer 2

$\Large \rm \bold \red {✰Question࿐}$

Prove that :-

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

$\Large \rm \bold \red {✰Solution࿐}$

$\green \mapsto \sf LHS = (sin\theta +cosec\theta)^2+(cos \theta +sec \theta)^2$

$\green \mapsto \sf sin^2\theta +cosec^2\theta+2sin \theta \: cosec \theta + {cos}^{2} \theta+sec^2\theta + 2cos \theta \: sec \theta$

$\green \mapsto \sf (sin^2\theta +cos^2\theta ) + 2 + 2 + cosec^2\theta + sec^2\theta$

$\green \mapsto \sf 1 + 2 + 2 + (1 + cot^2\theta)+(1 + tan^2\theta)$

$\green \mapsto \sf 1 + 2 + 2 + 1 + 1 + cot^2\theta +tan^2\theta$

$\green \mapsto \sf 7+tan^2\theta+cot^2\theta = RHS$

⭐ Hence, proved !!

$\Large \rm \bold \red {✰Concepts࿐}$

$\green \mapsto \sf sin^2\theta + {cos}^{2} \theta = 1$

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

$\green \mapsto \sf cosec^2\theta = 1 + cot^2\theta$

$\green \mapsto \sf sin \theta \: cosec \theta = 1$

$\green \mapsto \sf cos \theta \: sec \theta = 1$

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