Question

( 1 - tan theta ) square + ( 1 - cot theta) square = (sec theta - cosec theta) square​

Answer 1

Answer:

1+tanθ)2+(1+cotθ)2=(secθ+cscθ)2 , proved.

Step-by-step explanation:

To prove that, (1+\tan \theta)^2+(1+\cot \theta)^2=(\sec \theta+\csc\theta)^2(1+tanθ)2+(1+cotθ)2=(secθ+cscθ)2 .

L.H.S. = (1+\tan \theta)^2+(1+\cot \theta)^2(1+tanθ)2+(1+cotθ)2

= 1+\tan^2 \theta+2\tan \theta+1+\cot^2 \theta+2\cot \theta1+tan2θ+2tanθ+1+cot2θ+2cotθ

Using the algebraic identity,

(a+b)^{2}=a^{2}+b^{2}+2ab(a+b)2=a2+b2+2ab

= (1+\tan^2 \theta)+2\tan \theta+(1+\cot^2 \theta)+2\cot \theta(1+tan2θ)+2tanθ+(1+cot2θ)+2cotθ

Using the trigonometric identity,

\sec^2 \theta-\tan^2 \theta=1sec2θ−tan2θ=1

⇒ \sec^2 \theta=1+\tan^2 \thetasec2θ=1+tan2θ and

\csc^2 \theta-\cot^2 \theta=1csc2θ−cot2θ=1

⇒ \csc^2 \theta=1+\cot^2 \thetacsc2θ=1+cot2θ

= \sec^2 \theta+\csc^2 \theta++2(\tan \theta+\cot \theta)sec2θ+csc2θ++2(tanθ+cotθ)

= \sec^2 \theta+\csc^2 \theta+2(\dfrac{\sin \theta}{\cos \theta} +\dfrac{\cos \theta}{\sin \theta})sec2θ+csc2θ+2(cosθsinθ+sinθcosθ)

= \sec^2 \theta+\csc^2 \theta+2\dfrac{\sin^2 \theta+\cos^2 \theta}{\sin \theta\cos \theta}sec2θ+csc2θ+2sinθcosθsin2θ+cos2θ

Using the trigonometric identity,

\sin^2 \theta+\cos^2 \theta=1sin2θ+cos2θ=1

= \sec^2 \theta+\csc^2 \theta+2\dfrac{1}{\sin \theta\cos \theta}sec2θ+csc2θ+2sinθcosθ1

= \sec^2 \theta+\csc^2 \theta+2\sec \theta\csc \thetasec2θ+csc2θ+2secθcscθ

= (\sec \theta+\csc\theta)^2=(secθ+cscθ)2

= R.H.S., proved.

Thus, (1+\tan \theta)^2+(1+\cot \theta)^2=(\sec \theta+\csc\theta)^2(1+tanθ)2+(1+cotθ)2=(secθ+cscθ)2 , proved