Question

If
$\alpha = sin^{ - 1} (cos(sin^{ - 1}x ) \: and \: \beta = cos^{ - 1} (sin(cos^{ - 1} x))$
,then find
$tan \alpha .tan \beta .$
​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{\alpha=sin^{-1}\big\{cos\big(sin^{-1}(x)\big)\big\}\,\,\,\,,\,\,\,\,\beta=cos^{-1}\big\{sin\big(cos^{-1}(x)\big)\big\}}$

$\sf{\implies\alpha=sin^{-1}\left\{cos\left(cos^{-1}(\sqrt{1-x^2})\right)\right\}\,\,\,\,,\,\,\,\,\beta=cos^{-1}\left\{sin\left(sin^{-1}(\sqrt{1-x^2})\right)\right\}}$

$\sf{\implies\alpha=sin^{-1}\left(\sqrt{1-x^2}\right)\,\,\,\,,\,\,\,\,\beta=cos^{-1}\left(\sqrt{1-x^2}\right)}$

$\sf{\implies\,sin(\alpha)=\sqrt{1-x^2}\,\,\,\,,\,\,\,\,cos(\beta)=\sqrt{1-x^2}}$

$\sf{\implies\,cos(\alpha)=x\,\,\,\,,\,\,\,\,sin(\beta)=x}$

Now,

$\sf{tan(\alpha)}\cdot\sf{tan(\beta)}$

$=\sf{\dfrac{sin(\alpha)}{cos(\alpha)}}\cdot\sf{\dfrac{sin(\beta)}{cos(\beta)}}$

$=\sf{\dfrac{\sqrt{1-x^2}}{x}}\cdot\sf{\dfrac{x}{\sqrt{1-x^2}}}$

$\sf{=1}$