Question

prove that identity in trigonometry ​

Answer 1

Answer:

HEY mate here's your answer..

H.S. = \dfrac{cos\theta*cot\theta}{1+sin\theta}

1+sinθ

cosθ∗cotθ

= \dfrac{cos\theta * \dfrac{cos\theta}{sin\theta}*(1-sin\theta)}{(1+sin\theta)*(1-sin\theta)}

(1+sinθ)∗(1−sinθ)

cosθ∗

sinθ

cosθ

∗(1−sinθ)

[ by multiplying both the numerator and denominator by (1 - sinθ) ]

= \dfrac{\dfrac{cos^{2}\theta}{sin\theta}*(1-sin\theta)}{1-sin^{2}\theta}

1−sin

2

θ

sinθ

cos

2

θ

∗(1−sinθ)

= \dfrac{\dfrac{cos^{2}\theta}{sin\theta}*(1-sin\theta)}{cos^{2}\theta}

cos

2

θ

sinθ

cos

2

θ

∗(1−sinθ)

[ since sin^{2}\theta+cos^{2}\theta=1sin

2

θ+cos

2

θ=1 ]

= \dfrac{1-sin\theta}{sin\theta}

sinθ

1−sinθ

[ by cancelling cos^{2}\thetacos

2

θ from both the numerator and the denominator ]

= \dfrac{1}{sin\theta} - \dfrac{sin\theta}{sin\theta}

sinθ

1

−

sinθ

sinθ

(using the formula : \dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c})

c

a+b

=

c

a

+

c

b

)

= cosecθ - 1

= R.H.S.

Hence, proved.

Answer 2

Need to prove:

$\displaystyle\large\Longrightarrow\bf\dfrac{cos\theta cot\theta}{1+sin\theta}=cosec\theta-1$

Here we have to prove LHS=RHS.

So let's start!!

$\displaystyle{\Longrightarrow\large{\sf\dfrac{cos\theta cot\theta}{1+sin\theta}=cosec\theta-1}}$

Applying formula cot∅=cos∅/sin∅

$\displaystyle\large\Longrightarrow\sf\dfrac{cos\theta\times \dfrac{cos\theta}{sin\theta}}{1+sin\theta}=cosec\theta-1$

$\displaystyle\large\Longrightarrow\sf\dfrac{ \dfrac{cos^2\theta}{sin\theta}}{1+sin\theta}=cosec\theta-1$

$\displaystyle\large\Longrightarrow\sf\dfrac{cos^2\theta}{sin\theta(1+sin\theta)}=cosec\theta-1$

Now applying identity:

cos²∅=1-sin²∅

$\displaystyle\large\Longrightarrow\sf\dfrac{1-sin^2\theta}{\sin\theta(1+sin\theta)}=cosec\theta-1$

$\displaystyle\large\Longrightarrow\sf\dfrac{(1)^2-(sin\theta)^2}{\sin\theta(1+sin\theta)}=cosec\theta-1$

Now applying formula in numerator:

• A²-B²=(A+B)(A-B)

$\displaystyle\large\Longrightarrow\sf\dfrac{\big(1+sin\theta\big)\big(1-sin\theta\big)}{sin\theta(1+sin\theta)}=cosec\theta-1$

Now 1+sin∅ will be cancelled out from both numerator and denominator.

$\displaystyle\large\Longrightarrow\sf\dfrac{\big(1-sin\theta\big)}{sin\theta}=cosec\theta-1$

Now taking sin∅ on both numerators.

$\displaystyle\large\Longrightarrow\sf\dfrac{1}{sin\theta}-\dfrac{sin\theta}{sin\theta}=cosec\theta-1$

Now sin∅÷sin∅=1

$\displaystyle\large\Longrightarrow\sf\dfrac{1}{sin\theta}-1=cosec\theta-1$

Apply formula:

• 1/Sin∅=cosec∅

$\underline{\boxed{\red{\displaystyle\large\bf cosec\theta-1=cosec\theta-1}}}\star$

Hence proved LHS=RHS.

Trigonometric Identities:

$\boxed{\begin{minipage}{6cm} Important Trigonometric identities :- \\ \\ \: \: 1)\:\sin^2\theta+\cos^2\theta=1 \\ \\ 2)\:\sin^2\theta= 1-\cos^2\theta \\ \\ 3)\:\cos^2\theta=1-\sin^2\theta \\ \\ 4)\:1+\cot^2\theta=\text{cosec}^2 \, \theta \\ \\5)\: \text{cosec}^2 \, \theta-\cot^2\theta =1 \\ \\ 6)\:\text{cosec}^2 \, \theta= 1+\cot^2\theta \\\ \\ 7)\:\sec^2\theta=1+\tan^2\theta \\ \\ 8)\:\sec^2\theta-\tan^2\theta=1 \\ \\ 9)\:\tan^2\theta=\sec^2\theta-1 \end{minipage}}$