Question

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Answer 1

ANSWER:

GIVEN:

$\sqrt{sec {}^{2} \theta + cosec {}^{2} \theta }$

TO PROVE:

$\implies \sqrt{sec {}^{2} \theta + cosec {}^{2} \theta } = tan \theta + cot \theta$

FORMULA USED:

• 1+tan²Ø = Sec²Ø
• 1+cot²Ø = Cosec²Ø

SOLUTION:

$= \sqrt{sec {}^{2} \theta + cosec {}^{2} \theta } \\using \: the \: formula \\ = \sqrt{1 + tan {}^{2} \theta + 1 + cot {}^{2} \theta } \\ = \sqrt{tan {}^{2} \theta + cot {}^{2} \theta + 2 } \\ we \: know \: that \: (tan \theta \: \times \cot \theta \: = 1) \\ = \sqrt{tan {}^{2} \theta + cot {}^{2} \theta + 2 \times tan \theta \times \cot \theta } \\ = \sqrt{(tan \theta + cot \theta) {}^{2} } \\ = tan \theta \: + \cot \theta$

LHS = RHS

PROVED:

NOTE:

Some important formulas:

(a+b)² = a²+b²+2ab

(a-b)² = a²+b²-2ab

(a+b)(a-b) = a²-b²

(a+b)³ = a³+b³+3ab(a+b)

Answer 2

Answer:

Step-by-step explanation: L.H.S.

√(sec²theta+cosec²theta)

√(1/cos²theta+1/sin²theta)

√(sin²theta+cos²theta/sin²thetacos²theta)

√(1/sin²thetacos²theta)

1/sintheta costheta

R.H.S.

tan theta+cot theta

sin theta/cos theta+cos theta/sin theta

(cos²theta+sin²theta)/sin theta cos theta

1/sin theta cos theta

Proved

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