Question

tan(90+theta) + tan(90-theta)​

Answer 1

Answer:

To prove that: \dfrac{\tan \theta}{\tan (90- \theta)} + \dfrac{\sin (90 - \theta)}{\cos \theta} =\sec^2 \theta.

tan(90−θ)

tanθ

+

cosθ

sin(90−θ)

=sec

2

θ.

Solution:

L.H.S. = \dfrac{\tan \theta}{\tan (90- \theta)} + \dfrac{\sin (90 - \theta)}{\cos \theta}

tan(90−θ)

tanθ

+

cosθ

sin(90−θ)

Using the trigonometric identity:

\cot \thetacotθ = \tan (90- \theta)tan(90−θ) and

\cos \thetacosθ = \sin (90- \theta)sin(90−θ)

= \dfrac{\tan \theta}{\cot \theta} + \dfrac{\cos \theta}{\cos \theta}

cotθ

tanθ

+

cosθ

cosθ

= \dfrac{\tan \theta}{\cot \theta} + 1

cotθ

tanθ

+1

Using the trigonometric identity:

\tan \thetatanθ = \dfrac{1}{\cot \theta}

cotθ

1

= \dfrac{\tan \theta}{\dfrac{1}{\tan \theta} } + 1

tanθ

1

tanθ

+1

= \tan^2 \thetatan

2

θ + 1

Using the trigonometric identity:

\sec^2 \thetasec

2

θ - \tan^2 \thetatan

2

θ = 1

⇒ \sec^2 \thetasec

2

θ = 1 + \tan^2 \thetatan

2

θ

= \sec^2 \thetasec

2

θ

= R.H.S., proved.

Thus, \dfrac{\tan \theta}{\tan (90- \theta)} + \dfrac{\sin (90 - \theta)}{\cos \theta} =\sec^2 \theta

tan(90−θ)

tanθ

+

cosθ

sin(90−θ)

=sec

2

θ , proved.