Question

trigonometry maths
answer send in image​

Answer 1

cos^2 a/cos a-sin a is the 1st term

sin^2 a/sin a -cos a is the 2nd term

take -ve sign common from 2nd term it becomes

-sin^2 a/ cos a -sin a

now we add 2 terms as their denominators are same

so it will be

cos^2 a - sin ^2 a/cos a- sin a

now use the identity

a^2-b^2=(a-b)*(a+b)

we get

cos a +sin a

hence proved

if you liked it then mark me the brainest;)

have a nice day

Answer 2

AnswEr :

$\large\ : \implies\sf\red{\frac{cos\theta}{1 - tan\theta} + \frac{sin\theta}{1-cot\theta} }= \orange{sin\theta + cos\theta}$

$\underline{\dag\:\textsf{Taking \: left \: hand \: side \: of \: question:}}$

$\large\ : \implies\sf\red{\frac{cos\theta}{1 - tan\theta} + \frac{sin\theta}{1 - cot\theta} }$

$\large\ : \implies\sf\frac{cos\theta}{1 - \frac{sin\theta}{cos\theta} } + \frac{sin\theta}{1 - \frac{cos\theta}{sin\theta} }$

$\large\ : \implies\sf\frac{cos\theta}{ \frac{cos\theta - sin\theta}{cos\theta} } + \frac{sin\theta}{\frac{sin\theta - cos\theta}{cos\theta} }$

$\normalsize\ : \implies\sf\frac{cos^2\theta}{cos\theta - sin\theta} - \frac{sin^2\theta}{cos\theta - sin\theta} \\ \\ \normalsize\ : \implies\sf\frac{cos^2\theta - sin^2\theta} {cos\theta - sin\theta} \\ \\ \normalsize\ : \implies\sf\frac{\cancel{(cos\theta - sin\theta)}(cos\theta + sin\theta) }{\cancel{cos\theta - sin\theta}} \\ \\ \normalsize\ : \implies\sf\orange{ (cos\theta + sin\theta) } \\ \\ \large\ : \implies\sf\ L.H.S = R.H.S$

Some important related to it :

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\cos^2\theta=1-\sin^2\theta \\ \\ 4)1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5) \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\sec^2\theta=1+\tan^2\theta \\ \\ 8)\sec^2\theta-\tan^2\thetha=1 \\ \\ 9)\tan^2\theta=\sec^2\theta-1 \end{minipage}}$