Question

plzzz help..... me ​

Answer 1

Given:-

tanθ = x .

To Find:-

The value of secθ.

Formula Used:-

We will use a identity ,

$\large\pink{\underline{\boxed{\purple{\tt{\dag \:\:sec^2\theta\:\:-\:\:tan^2\theta\:\:=\:\:1\:\: }}}}}$

Solution:-

Given equⁿ is : tanθ = x .

Now , take the given identity :

$\tt:\implies sec^2\theta -\tan^2\theta=1$

$\tt:\implies sec^2\theta = 1 + \tan^2\theta$

$\tt:\implies sec\theta=\sqrt{1+\tan^2\theta}$

$\tt:\implies \sec\theta = \sqrt{1+(x)^2}$

$\underline{\boxed{\red{\tt \longmapsto \sec\theta \:\:=\:\:\sqrt{1+x^2}}}}$

More to know :-

Here are some more formula .

$\green{\boxed{\red{\bullet \tt sin^2\theta+cos^2\theta=1}}}$

$\orange{\boxed{\red{\bullet \tt cosec^2\theta-cot^2\theta=1}}}$

$\pink{\boxed{\red{\bullet \tt sin(A+B)=sinA.cosB + cosA.sinB }}}$

$\blue{\boxed{\red{\bullet \tt sin(A-B)=sinA.cosB - cosA.sinB}}}$

$\purple{\boxed{\red{\bullet \tt cos(A+B)=cosA.cosB - sinA.sinB}}}$

$\blue{\boxed{\red{\bullet \tt cos(A-B)=cosA.cosB + sinA.sinB}}}$

$\blue{\boxed{\red{\bullet \tt tan(A+B)=\dfrac{tanA+tanB}{1-tanA.tanB} }}}$

$\green{\boxed{\red{\bullet \tt tan(A-B)=\dfrac{tanA-tanB}{1+tanA.tanB} }}}$

Answer 2

Answer:

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