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ratelena [41]
3 years ago
10

Prove the following identities : i) tan a + cot a = cosec a sec a​

Mathematics
1 answer:
inysia [295]3 years ago
7 0

Step-by-step explanation:

\tan \alpha + \cot\alpha = \dfrac{\sin \alpha}{\cos \alpha} +\dfrac{\cos \alpha}{\sin \alpha}

=\dfrac{\sin^2\alpha + \cos^2\alpha}{\sin\alpha\cos\alpha}=\dfrac{1}{\sin\alpha\cos\alpha}

=\left(\dfrac{1}{\sin\alpha}\right)\!\left(\dfrac{1}{\cos\alpha}\right)=\csc \alpha \sec\alpha

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Multiply 20 ounces by .20 (The percent) then add your answer to the 20 ounces.

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pshichka [43]

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A carnival ride is in the shape of a wheel with a radius of 15 feet. The wheel has 24 cars attached to the center of the wheel.
Anarel [89]
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3 0
2 years ago
Cot? 2x + cos2x + sin’ 2x = csc? 2x<br> Need help with 40
kkurt [141]

Answer:

\rule{100mm}{0.1mm}

    \cot^2(2x)+\cos^2(2x)+\sin^2(2x)=\csc^2(2x)

\dfrac{\cos^2(2x)}{\sin^2(2x)} +[(\cos^2(2x)+\sin^2(2x)]=\csc^2(2x)

                               \dfrac{\cos^2(2x)}{\sin^2(2x)} +1=\csc^2(2x)

                    \dfrac{\cos^2(2x)+\sin^2(2x)}{\sin^2(2x)}=\csc^2(2x)

                                    \dfrac{1}{sin^2(2x)} =\csc^2(2x)

                                     \boxed{\csc^2(2x)=\csc^2(2x)} ✔

\rule{100mm}{0.1mm}

4 0
2 years ago
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