Question

Please these questions

With explanation

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Question 3 -

      Solution : -

$\implies cot\theta - tan\theta \\\\\implies \dfrac{cos\theta}{sin\theta} - \dfrac{sin\theta}{cos\theta}\\\\\\\implies \dfrac{cos^2 \theta - sin^2 \theta}{sin\theta cos \theta}\\\\\\\implies \dfrac{cos^2 \theta-(1-cos^2 \theta)}{sin\theta cos\theta}\quad\quad\bold{\bigg\{sin^2 \theta = 1 - cos^2 \theta}\\\\\\\implies \dfrac{cos^2 \theta - 1 + cos^2 \theta}{sin\theta cos\theta}\\\\\\\implies \dfrac{2cos^2 \theta -1 }{sin\theta cos\theta}$

Proved.

Question 4 -

     Solution : -

$\implies sin\theta ( 1 + tan\theta ) + cos\theta(1 + cot\theta) \\\\\implies sin \theta ( 1 + tan\theta ) + cos\theta \bigg( 1 + \dfrac{1}{tan\theta}\bigg) \\\\\\\implies sin\theta( 1 + tan\theta) + cos\theta \bigg(\dfrac{tan\theta +1 }{tan\theta}\bigg)\\\\\\\implies ( 1+tan\theta) \bigg[sin\theta( 1) +\dfrac{cos\theta}{tan\theta}\bigg]\\\\\\\implies ( 1 + tan\theta)\bigg\{ sin\theta + \dfrac{cos\theta}{\frac{sin\theta}{cos\theta}}\bigg\}$

$\implies ( 1 + tan\theta)\bigg\{sin\theta + \dfrac{cos^2 \theta}{sin \theta}\bigg\}\\\\\\\implies (1 + tan\theta)\bigg\{ \dfrac{sin^2 \theta+cos^2 \theta}{sin\theta}\bigg\}\\\\\\\implies \dfrac{ 1 + tan\theta}{sin\theta}\quad\quad\bold{\bigg\{sin^2\theta + cos^2 \theta = 1 }\\\\\\\implies cosec\theta + \dfrac{1}{cos\theta}\\\\\\\implies cosec\theta+ sec\theta$

Proved.

Question 5 -

     Solution : -

Note : Theta is written as A.

Given,secA + tanA = p   ...( i )

We know, sec^2 A - tan^2 A or ( secA + tanA )( secA - tanA ) = 1

Substitute the value of ( secA + tanA ) :

⇒ ( p )( secA - tanA ) = 1

⇒ secA - tanA = 1 / p    ...( ii )

On subtracting ( ii ) from ( i ), we get

⇒ 2 tanA = p - 1 / p

$\implies tanA = \dfrac{p^2 - 1 }{ 2p }\\\\\\\implies \dfrac{1}{cotA} = \dfrac{p^2 - 1}{2p}\\\\\\\implies cotA = \dfrac{2p}{p^2 - 1}\\\\\\\implies cot^2 A = \bigg(\dfrac{2p}{p^2 - 1} \bigg)^2 \quad\quad\mathsf{\bigg\{Square\:on\:both\:sides}\\\\\\\implies 1 + cot^2A = \dfrac{4p^2}{(p^2-1)^2}+1\quad\quad\mathsf{\bigg\{Adding\:1\:on\:both\:sides}\\\\\\\implies cosec^2 A = \dfrac{4p^2 + p^4 +1-2p^2}{(p^2-1)^2}\quad\quad\bigg\{ (p^2-1)^2 = p^4+1-2p^2\quad 1 + cot^2A = cosec^2 A$

$\implies cosec^2 A = \dfrac{p^4+1+2p^2}{(p^2-1)^2}\\\\\\\implies cosec^2 A = \dfrac{(p^2+1)^2}{(p^2-1)^2}\\\\\\\implies cosec A = \dfrac{p^2+1}{p^2-1}\:\:or\:\:-\bigg(\dfrac{p^2+1}{p^2 -1}\bigg)$

Hence,

Value of cosecθ is ( p^2 + 1 )/(p^2 - 1 )  or  - (p^2+1)/(p^2-1)