Question

sin square minus 160 degree upon sin square 70degree + sin 180 - theta upon sin theta is equals to sec square 20 degree​

Answer 1

Answer:

Proved.

Step-by-step explanation:

We have to prove that,

$\frac{Sin^{2}(-160)}{Sin^{2}70 } +\frac{Sin(180-\alpha) }{Sin\alpha }=Sec^{2} 20$...... (1)

Now the left hand side of equatio (1) = $\frac{Sin^{2}(-160)}{Sin^{2}70 } +\frac{Sin(180-\alpha) }{Sin\alpha }$

= $\frac{Sin^{2}(160)}{Sin^{2}(90-20) } +\frac{Sin(180-\alpha) }{Sin\alpha }$

{Since $Sin^{2} (-\alpha )=(-Sin\alpha) ^{2}= Sin^{2}\alpha$}

= $\frac{Sin^{2}(180-20) }{Cos^{2}20 }+\frac{Sin\alpha }{Sin\alpha }$

= $\frac{Sin^{2}20 }{Cos^{2}20 } +1$

{ Where we know that, $Sin(90-\alpha )=Cos\alpha$ and $Sin(180-\alpha )=Sin\alpha$}

= $tan^{2} 20+1$ { Apply the formula, $Sec^{2} \alpha -tan^{2} \alpha =1$}

= $Sec^{2} 20$

= Right hand side of equation (1)

Hence, proved.

Answer 2

Answer:

Answer:

Proved.

Step-by-step explanation:

We have to prove that,

\frac{Sin^{2}(-160)}{Sin^{2}70 } +\frac{Sin(180-\alpha) }{Sin\alpha }=Sec^{2} 20

Sin

2

70

Sin

2

(−160)

+

Sinα

Sin(180−α)

=Sec

2

20 ...... (1)

Now the left hand side of equatio (1) = \frac{Sin^{2}(-160)}{Sin^{2}70 } +\frac{Sin(180-\alpha) }{Sin\alpha }

Sin

2

70

Sin

2

(−160)

+

Sinα

Sin(180−α)

= \frac{Sin^{2}(160)}{Sin^{2}(90-20) } +\frac{Sin(180-\alpha) }{Sin\alpha }

Sin

2

(90−20)

Sin

2

(160)

+

Sinα

Sin(180−α)

{Since Sin^{2} (-\alpha )=(-Sin\alpha) ^{2}= Sin^{2}\alphaSin

2

(−α)=(−Sinα)

2

=Sin

2

α }

= \frac{Sin^{2}(180-20) }{Cos^{2}20 }+\frac{Sin\alpha }{Sin\alpha }

Cos

2

20

Sin

2

(180−20)

+

Sinα

Sinα

= \frac{Sin^{2}20 }{Cos^{2}20 } +1

Cos

2

20

Sin

2

20

+1

{ Where we know that, Sin(90-\alpha )=Cos\alphaSin(90−α)=Cosα and Sin(180-\alpha )=Sin\alphaSin(180−α)=Sinα }

= tan^{2} 20+1tan

2

20+1 { Apply the formula, Sec^{2} \alpha -tan^{2} \alpha =1Sec

2

α−tan

2

α=1 }

= Sec^{2} 20Sec

2

20

= Right hand side of equation (1)

Hence, proved.