Question

1+tan theta*tan theta/2=sectheta​

Answer 1

Let us know some trigonometric formulae to solve the problem:

• $\mathsf{tanA=\dfrac{2\:tan\dfrac{A}{2}}{1-tan^{2}\dfrac{A}{2}}}$

• $\mathsf{cosA=\dfrac{1-tan^{2}\dfrac{A}{2}}{1+tan^{2}\dfrac{A}{2}}}$

Proof of the problem:

Now, $\mathsf{L.H.S=1+tan\theta\:tan\dfrac{\theta}{2}}$

$\mathsf{=1+\dfrac{2\:tan\dfrac{\theta}{2}}{1-tan^{2}\dfrac{\theta}{2}}\:tan\dfrac{\theta}{2}}$

$\mathsf{=1+\dfrac{2\:tan^{2}\dfrac{\theta}{2}}{1-tan^{2}\dfrac{\theta}{2}}}$

$\mathsf{=\dfrac{1-tan^{2}\dfrac{\theta}{2}+2\:tan^{2}\dfrac{\theta}{2}}{1-tan^{2}\dfrac{\theta}{2}}}$

$\mathsf{=\dfrac{1+tan^{2}\dfrac{\theta}{2}}{1-tan^{2}\dfrac{\theta}{2}}}$

$\mathsf{=\dfrac{1}{\dfrac{1-tan^{2}\dfrac{\theta}{2}}{1+tan^{2}\dfrac{\theta}{2}}}}$

$\mathsf{=\dfrac{1}{cos\theta}}$

$\mathsf{=sec\theta=R.H.S.}$

Therefore, $\boxed{\mathsf{1+tan\theta\:tan\dfrac{\theta}{2}=secA}}$ (proved)

Answer 2

Answer:

oneplus 2 tan theta / 2 upon 1 - 10 square theta / 2 * 10 theta by 2 is equal to 1 + 210 square theta upon 2 / 1 - 10 square theta / 2 is equal to 1 - 10 square theta by 2 + 2 tan square theta / 2 / 1 - 10 square theta by 2 is equal to 1 + 10 square theta / 2 upon 1 - 10 squared theta by 2 is equal to 1 upon 1 - 2 upon 1 + 10 square theta / 2 is equal to 1 upon cos theta is equal to RHS