प्रतिलोम त्रिकोणमिति Important Questions | Class 12th Maths | Full Concepts & Solutions

Описание: 1. Class 12 Maths | Inverse Trigonometry Important Questions with Solutions | Board Exam Special


2. प्रतिलोम त्रिकोणमिति Important Questions | Class 12th Maths | Full Concepts & Solutions


3. Inverse Trigonometric Functions – Top Questions for Boards | Full Explanation Hindi


4. प्रतिलोम त्रिकोणमिति | 1 वीडियो में पूरा चैप्टर + महत्वपूर्ण प्रश्न समाधान

📌 YouTube Description (Hindi + English Mix)

Description:
इस वीडियो में हम प्रतिलोम त्रिकोणमिति (Inverse Trigonometric Functions) के सभी महत्वपूर्ण प्रश्नों को आसान भाषा में समझेंगे।
यह वीडियो Class 12th Board Exams, JEE, NEET, और Competitive Exams के लिए बेहद उपयोगी है।
वीडियो में शामिल हैं—
✔ Domain & Range
✔ Principal Values
✔ Important Identities
✔ NCERT + Board Pattern Questions
✔ Previous Year Questions (PYQ)
✔ Step-by-step Solutions

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🎓 Class 12th Maths के लिए यह वीडियो जरूर देखें।

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📌 Important Questions with Solutions (Short Notes Form)
Q1. Find the value of

\sin^{-1}\left(\frac{1}{2}\right)

Solution:

\sin^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}--

Q2. Find the value of

\cos^{-1}\left(-\frac{1}{2}\right)

Solution:

Principal value of lies in

\cos^{-1}\left(-\frac{1}{2}\right)=\frac{2\pi}{3}


Q3. Simplify:

\tan^{-1}(1)+\tan^{-1}(2)

Solution:

Using identity:

\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)

Here,

a=1, b=2

\tan^{-1}(1)+\tan^{-1}(2)=\tan^{-1}\left(\frac{3}{1-2}\right)=\tan^{-1}(-3)

=\ -\tan^{-1}(3)


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Q4. Find the value of:

\sin^{-1}(x)+\cos^{-1}(x)

Solution:

Identity:

\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2}


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Q5. Solve:

\cos^{-1}\left(\frac{2}{\sqrt{5}}\right) - \sin^{-1}\left(\frac{1}{\sqrt{5}}\right)

Solution:

Use identity:

\cos^{-1}(x) = \sin^{-1}(\sqrt{1-x^2})

\cos^{-1}\left(\frac{2}{\sqrt{5}}\right) = \sin^{-1}\left(\frac{1}{\sqrt{5}}\right)

So,

\boxed{0}


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Q6. Evaluate:

\tan^{-1}\left(\frac{3}{4}\right)+\cot^{-1}\left(\frac{4}{3}\right)

Solution:

\tan^{-1}(x)+\cot^{-1}(x)=\frac{\pi}{2}

\boxed{\frac{\pi}{2}}


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Q7. Prove:

\sin^{-1}(x) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)

Proof:

Let

\sin^{-1}(x)=\theta \Rightarrow x=\sin\theta

\cos\theta=\sqrt{1-x^2} 

\tan\theta=\frac{x}{\sqrt{1-x^2}}

So,

\theta = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)

Hence proved ✔