Question

tan%7D%5E%7B2%7Da%20%20%2B%20%20%7Bcot%7D%5E%7B2%7D%20a%20%2B%202%20%5C%5C%20" id="TexFormula1" title="sec {}^{2} a \: . \: cosec {}^{2} a = {tan}^{2}a + {cot}^{2} a + 2 \\ " alt="sec {}^{2} a \: . \: cosec {}^{2} a = {tan}^{2}a + {cot}^{2} a + 2 \\ " align="absmiddle" class="latex-formula">
Please help!!!!!!!!!!!!​

Answer 1

Answer:

Trigonometric identities

$\sec^2(\alpha)=1+\tan^2(\alpha)$

$\csc^2(\alpha)=1+\cot^2(\alpha)$

$\cot^2(\alpha)=\dfrac{1}{\tan^2(\alpha)}$

Solution

$\begin{aligned}\sec^2(\alpha) \cdot \csc^2(\alpha) & = (1+\tan^2(\alpha))(1+\cot^2(\alpha))\\\\ & =1+\cot^2(\alpha)+\tan^2(\alpha)+tan^2(\alpha)\cot^2(\alpha)\\\\ & = 1+\cot^2(\alpha)+\tan^2(\alpha)+\tan^2(\alpha) \cdot \dfrac{1}{\tan^2(\alpha)}\\\\ & = 1+\cot^2(\alpha)+\tan^2(\alpha)+ \dfrac{\tan^2(\alpha)}{\tan^2(\alpha)}\\\\& = 1+\cot^2(\alpha)+\tan^2(\alpha)+1\\\\& = \tan^2(\alpha)+\cot^2(\alpha)+2\end{aligned}$

Answer 2

Answer:

See below ~

Step-by-step explanation:

Identities Needed

1. sec a = 1 / cos a
2. cosec a = 1 / sin a
3. cot a = 1 / tan a

Proving

• (sec²a)(cosec²a) = tan²a + cot²a + 2

Taking the LHS

• sec²a x cosec²a
• 1/cos²a x 1/sin²a
• 1/(sin²a)(cos²a) -(Equation 1)

Taking the RHS

• tan²a + cot²a + 2
• tan²a + cot²a + 2(tana)(cota)
• (tana + cota)²
• (sina/cosa + cosa/sina)²
• (sin²a + cos²a / sinacosa)²
• (1/sinacosa)²
• 1/(sin²a)(cos²a)  -(Equation 2)

∴ Equation 1 = Equation 2

∴ LHS = RHS

Hence, proved.