Maths Questions

Solution : In right angled triangle ABC, \(sin \theta\) = \(BC\over AC\)  \(\implies\)   \(sin^2 \theta\) = \(BC^2\over AC^2\) \(cos \theta\) = \(AB\over AC\)  \(\implies\)   \(cos^2 \theta\) = \(AB^2\over AC^2\) On adding, \(sin^2 \theta\) + \(cos^2 \theta\) = \(BC^2\over AC^2\) + \(AB^2\over AC^2\) \(sin^2 \theta\) + \(cos^2 \theta\) = \(BC^2 + AB^2\over AC^2\) = \(AC^2\over AC^2\) […]

Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\),  then \(\angle\) C = 90 – \(\theta\) sec A = \(sec\theta\) = \(AC\over AB\)                     ……..(1) cosec C = \(cosec(90 – \theta)\) = \(AC\over AB\)           

Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\),  then \(\angle\) C = 90 – \(\theta\) cosec A = \(cosec\theta\) = \(AC\over BC\)                     ……..(1) sec C = \(sec(90 – \theta)\) = \(AC\over BC\)           

Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\),  then \(\angle\) C = 90 – \(\theta\) tan A = \(tan\theta\) = \(BC\over AB\)                     ……..(1) cot C = \(cot(90 – \theta)\) = \(BC\over AB\)           

Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\),  then \(\angle\) C = 90 – \(\theta\) cot A = \(cot\theta\) = \(AB\over BC\)                     ……..(1) tan C = \(tan(90 – \theta)\) = \(AB\over BC\)           

Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\),  then \(\angle\) C = 90 – \(\theta\) sin A = \(sin\theta\) = \(BC\over AC\)                     ……..(1) cos C = \(cos(90 – \theta)\) = \(BC\over AC\)           

Solution : Draw a right angled triangle ABC right angled at B. Let \(\angle\) A = \(\theta\),  then \(\angle\) C = 90 – \(\theta\) cos A = \(cos\theta\) = \(AB\over AC\)                     ……..(1) sin C = \(sin(90 – \theta)\) = \(AB\over AC\)           

Solution : The value of cosec 90 degrees is 1. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point B. When \(\angle\)

Solution : The value of sec 90 degrees is \(\infty\) or not defined. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point

Solution : The value of tan 90 degrees is \(\infty\) or not defined. Proof : \(\angle\) A of \(\Delta\) ABC is made large and large until it becomes 90 degrees. As \(\angle\) A gets large and large \(\angle\) C gets smaller and smaller. Side AB goes on decreasing. The point A gets closer to point B.