Question

Solve the attached question.NO spam!Step by step explanation needed!​

Answer 1

LHS= $\tan {}^{2} φ + \cot {}^{2} φ + 2$

$= \frac{ \sin^{2}φ }{ \cos {}^{2}φ } + \frac{ \cos {}^{2} φ }{ \sin {}^{2} φ} + 2$

$= \frac{ \sin {}^{4}φ + \cos {}^{4} φ + 2 \sin {}^{2} φ \cos {}^{2} φ }{ \sin {}^{2}φ. \cos {}^{2} φ }$

$= \frac{( \sin {}^{2}φ + \cos {}^{2} φ ) {}^{2} }{ \sin {}^{2} φ. \cos {}^{2} φ } = \frac{1}{ \sin {}^{2} φ} \times \frac{1}{ \cos {}^{2} φ}$

$= \sec {}^{2} φ \cosec {}^{2} φ$

= RHS

Hence, proved✔︎

Answer 2

Answer:

Step-by-step explanation:

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