Question

prove that tan theta + sec theta minus 1 upon 10 theta minus sec theta + 1 is equal to 1 + sin theta upon cos theta

Answer 1

$\underline{\textsf{To prove:}}$

$\mathsf{\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}=\dfrac{1+sin\theta}{cos\theta}}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}}$

$\textsf{Using the identity,}$

$\mathsf{sec^2A-tan^2A=1}$

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

$\mathsf{=\dfrac{sec\theta+tan\theta-(sec\theta-tan\theta)(sec\theta+tan\theta)}{tan\theta-sec\theta+1}}$

$\mathsf{=\dfrac{(sec\theta+tan\theta)[1-(sec\theta-tan\theta)]}{tan\theta-sec\theta+1}}$

$\mathsf{=\dfrac{(sec\theta+tan\theta)[1-sec\theta+tan\theta]}{1-sec\theta+tan\theta}}$

$\mathsf{=sec\theta+tan\theta}$

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

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

$\implies\boxed{\mathsf{\dfrac{tan\theta+sec\theta-1}{tan\theta-sec\theta+1}=\dfrac{1+sin\theta}{cos\theta}}}$

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Answer 2

Step-by-step explanation:

answer is above in the picture