Question

plzz answer my question
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step-by-step ​

Answer 1

Answer:

see the attached image

hope it helps

Answer 2

Answer:

The value of

$\boxed{\red{\sf\:\sec^{2}\:\theta\:-\:\tan^{2}\:\theta\:=\:1}}$

Step-by-step-explanation:

NOTE: Kindly refer to the attachment first.

In right angled $\sf\:\triangle\:PQR$ as shown in figure,

$\sf\:\angle\:PQR\:=\:90^{\circ}\:and\:\angle\:PRQ\:=\:\theta$

Here,

$\sf\:sin\:\theta\:=\:\dfrac{Opposite\:side}{Hypotenuse}\:=\:\frac{PQ}{PR}\\\\\sf\:cos\:\theta\:=\:\dfrac{Adjacent\:side}{Hypotenuse}\:=\:\frac{QR}{PR}\\\\\sf\:tan\:\theta\:=\:\dfrac{Opposite\:side}{Adjacent\:side}\:=\:\frac{PQ}{QR}\\\\\sf\:cosec\:\theta\:=\:\dfrac{Hypotenuse}{Opposite\:side}\:=\:\frac{PR}{PQ}\\\\\sf\:sec\:\theta\:=\:\dfrac{Hypotenuse}{Adjacent\:side}\:=\:\frac{PR}{QR}\\\\\sf\:cot\:\theta\:=\:\dfrac{Adjacent\:side}{Opposite\:side}\:=\:\frac{QR}{PQ}$

Now,

$\boxed{\pink{\sf\:PR^{2}\:=\:PQ^{2}\:+\:QR^{2}}}\:\:\:-\:[\:\sf\:Pythagoras\:theorem\:]\:\:-\:(\:1\:)$

Dividing both sides of ( 1 ) by $\sf\:PR^{2}$, we get,

$\sf\:\frac{\cancel{PR^{2}}}{\cancel{PR^{2}}}\:=\:\frac{PQ^{2}}{PR^{2}}\:+\:\frac{QR^{2}}{PR^{2}}\\\\\implies\sf\:1\:=\:(\:\frac{PQ}{PR}\:)^{2}\:+\:(\:\frac{QR}{PR}\:)^{2}\\\\\implies\sf\:1\:=\:(\:sin\:\theta\:)^{2}\:+\:(\:cos\:\theta\:)^{2}\\\\\implies\boxed{\red{\sf\:sin^{2}\:\theta\:+\:cos^{2}\:\theta\:=\:1}}\:\:\:-\:-\:-(\:1\:)$

Dividing each term of ( 2 ), by $\sf\:cos^{2}\:\theta$, we get,

$\sf\:\frac{sin^{2}\:\theta}{cos^{2}\:\theta}\:+\:\frac{cos^{2}\:\theta}{cos^{2}\:\theta}\:=\:\frac{1}{cos^{2}\:\theta}\\\\\implies\sf\:tan^{2}\:\theta\:+\:1\:=\:sec^{2}\:\theta\:\:[\:\because\:\frac{sin\:\theta}{cos\:\theta}\:=\:\tan\:\theta\:and\:\frac{1}{cos\:\theta}\:=\:sec\:\theta\:]\\\\\implies\boxed{\red{\sf\:sec^{2}\:\theta\:-\:tan^{2}\:\theta\:=\:1}}$

$\boxed{\begin{minipage}{7cm} Fundamental\:Trigonometric\:Identities\\\\ \sin^{2}\:\theta\:+\:\cos^{2}\:\theta=1\\\\1+\tan^{2}\:\theta\:=\:\sec^{2}\:\theta\\\\1+\cot^{2}\:\theta=\text{cosec}^{2}\:\theta \end {minipage}}$