Question

Hola Brainlians. Here's a question to check your Trigonometric skill. ICSE Board : Inverse Trigonometry, Class 10. Give answer fast and no spam.

Proof the given identity.

[cos x / (1 + sin x)] + [(1 + sin x)/cos x] = sec A

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Answer 1

★ Concept :-

Here the concept of Trigonometric Identities have been used. We see that we are given an equation where we need to prove L.H.S. and R.H.S. to be equal. Firstly we can take L.C.M. of the fractions and then solve it. Then we can substitute different formulas there anf prove it to be equal to R.H.S.

Let's do it !!

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★ Solution :-

Given to prove,

$\;\;\bf{\mapsto\;\;\green{\bigg[\dfrac{\cos x}{(1\:+\:\sin x)}\bigg]\:+\:\bigg[\dfrac{(1\:+\:\sin x)}{\cos x}\bigg]\;=\;2\sec x}}$

This is the appropriate qúestion.

From here, we get

$\;\;\sf{\odot\;\;\blue{L.H.S.\;=\;\bigg[\dfrac{\cos x}{(1\:+\:\sin x)}\bigg]\:+\:\bigg[\dfrac{(1\:+\:\sin x)}{\cos x}\bigg]}}$

And,

$\;\;\sf{\odot\;\;\orange{R.H.S.\;=\;2\sec x}}$

Now let's work upon L.H.S. Then,

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\bigg[\dfrac{\cos x}{(1\:+\:\sin x)}\bigg]\:+\:\bigg[\dfrac{(1\:+\:\sin x)}{\cos x}\bigg]}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{\cos x}{(1\:+\:\sin x)}\:+\:\dfrac{(1\:+\:\sin x)}{\cos x}}$

Now let's take L.C.M. of the denominators.

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(\cos x)(\cos x)}{(1\:+\:\sin x)(\cos x)}\:+\:\dfrac{(1\:+\:\sin x)(1\:+\:\sin x)}{(\cos x)(1\:+\:\sin x)}}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{(\cos x)(\cos x)\:+\:(1\:+\:\sin x)(1\:+\:\sin x)}{(1\:+\:\sin x)(\cos x)}}$

By combining two fractions we got it.

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{\cos^{2}x\:+\:(\sin^{2}x\:+\:1\:+\:2\sin x)}{(1\:+\:\sin x)(\cos x)}}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{\cos^{2}x\:+\:\sin^{2}x\:+\:1\:+\:2\sin x}{(1\:+\:\sin x)(\cos x)}}$

• sin² A + cos² A = 1 (This is the formula)

• Here A = x

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{1\:+\:1\:+\:2\sin x}{(1\:+\:\sin x)(\cos x)}}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{2\:+\:2\sin x}{(1\:+\:\sin x)(\cos x)}}$

Now taking 2 as common from Numerator, we get

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{2(1\:+\:\sin x)}{(1\:+\:\sin x)(\cos x)}}$

Now cancelling the like terms that is (1 + sin x) from both numerator and denominator we get,

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;\dfrac{2}{(\cos x)}}$

$\;\;\tt{\rightarrow\;\;L.H.S.\;=\;2\bigg(\dfrac{1}{\cos x}\bigg)}$

• sec A = 1/cos A (This is the formula)

• Here A = x

$\;\;\tt{\odot\;\;L.H.S.\;=\;2\bigg(\sec x\bigg)}$

$\;\;\bf{\rightarrow\;\;\red{L.H.S.\;=\;2\sec x}}$

This is the answer.

Now we can include R.H.S. also here.

$\;\;\bf{\rightarrow\;\;\purple{L.H.S.\;=\;R.H.S.\;=\;\;2\sec x}}$

Hence, Proved.

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★ More to know :-

• sec² A = 1 + tan² A

• cosec² A = 1 + cot² A

• cosec A = 1/sin A

• sec A = 1/cos A

• cot A = 1/tan A

Answer 2

Answer:

Answer in the attachment.