Question

Cos²20+cos²70+sin48sec42+cos40 cosec50

Answer 1

$\bf\large\color{blue}\mathcal{HEYA\:\:FRIEND\:\:}\mathbb{:)}\\\bf\large\color{violet}\textit{Here\:is\:your\:answer}$

Using,
$\sin^{2} \: + \: \cos^{2} \: = \: 1$ ___(1)
And

• Sin(90-A) = CosA __(2)

• Cos(90-A) = SinA __(3)

*{Taking A as theta}

And
$\sec = \frac{1}{ \cos }$ _____(4)
$\cosec \: = \frac{1}{sin}$ _____(5)
____________________________________

$\cos^{2}20 + cos^{2} 70 + sin48.sec42 + \\cos40. cosec50$
____________________________________

$\cos^{2}20 + cos^{2}(90-70)\\+sin(90-48).sec42 \\+ cos(90-40).cosec50$.
{from (2) and (3)}
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$\cos^{2}20 + sin^{2}20 +cos42.sec42\\ + sin50.cosec50$
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$1 + cos42.\frac{1}{cos}+ sin50.\frac{1}{sin50}$
{from (1)}
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1 + 1 + 1 = 3
____________________________________

Hope this helps!!!!!

Answer 2

Step-by-step explanation:

hope this helps you guys !!!!