Question

the value of Tan² alpha-1/cos²alpha​

Answer 1

Step-by-step explanation:

i hope it may help you......

Answer 2

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To FinD :

$\sf \rightsquigarrow { \tan}^{2} \alpha - \frac{1}{ { \cos}^{2} \alpha } = {?}$

SolutioN :

$\circ \: \: \: \sf \: we \: \: know \: \: the \: \: formula,$

$\hookrightarrow \sf \: { \sec}^{2} \theta - { \tan}^{2} \theta = 1$

Now,

$\: \: \: \: \: \: \: \: \: \: \: \sf \: { \tan}^{2} \alpha - \frac{1}{ { \cos}^{2} \alpha }$

$\sf \implies- ( - { \tan}^{2} \alpha + { \sec}^{2} \alpha )$

$\implies{ \underline {\boxed{ \: - \sf1 \: }}}$

ConcepT BoosteR :

$\circ \: \: { \sin}^{2} \theta + { \cos}^{2} \theta = 1$

$\circ \: \: { \cosec}^{2} \theta + { \cot}^{2} \theta = 1$

$\circ \sf \: \: \sin( \theta + \phi) = \sin \theta \cos \phi + \cos \theta \sin \phi$

$\circ \sf \: \: \sin( \theta - \phi) = \sin \theta \cos \phi - \cos \theta \sin \phi$

$\circ \sf \: \: \cos( \theta + \phi) = \cos \theta \cos \phi - \sin \theta \sin \phi$

$\circ \sf \: \: \cos( \theta - \phi) = \cos \theta \cos \phi + \sin \theta \sin \phi$

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HOPE THIS IS HELPFUL...

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