Question

prove that (sin 0 + sec 0)² + (cos 0 + sec 0)²=7 tan²0+cot²0​

Answer 1

Answer:

Answer with Step-by-step explanation:

LHS

(sin\theta+cosec\theta)^2+(cos\theta+sec\theta)^2(sinθ+cosecθ)

2

+(cosθ+secθ)

2

sin^2\theta+cosec^2\theta+2sin\theta cosec\theta+sec^2\theta+cos^2\theta+2cos\theta sec\thetasin

2

θ+cosec

2

θ+2sinθcosecθ+sec

2

θ+cos

2

θ+2cosθsecθ

Using identity :(a+b)^2=a^2+b^2+2ab(a+b)

2

=a

2

+b

2

+2ab

We know that

tan^2\theta+1=sec^2\theta, 1+cot^2\theta=cosec^2\theta,sec\theta=\frac{1}{cos\theta},cosec\theta=\frac{1}{sin\theta}tan

2

θ+1=sec

2

θ,1+cot

2

θ=cosec

2

θ,secθ=

cosθ

1

,cosecθ=

sinθ

1

sin^2\theta+cos^2\theta=1sin

2

θ+cos

2

θ=1

Using the formula

1+1+cot^2\theta+2sin\theta\times \frac{1}{sin\theta}+1+tan^2\theta+2cos\theta\times \frac{1}{cos\theta}1+1+cot

2

θ+2sinθ×

sinθ

1

+1+tan

2

θ+2cosθ×

cosθ

1

3+tan^2\theta+cot^2\theta+2+23+tan

2

θ+cot

2

θ+2+2

7+tan^2\theta+cot^2\theta7+tan

2

θ+cot

2

θ

LHS=RHS

Hence, proved.

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