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New batch starting from 27 Jan 2016 for English at 11 am. Special class by Sagar sir for competitive English.

Important TRIGONOMETRIC ratios.
We know that,
sin (90° + θ) = cos θ
cos (90° + θ) = - sin θ
tan (90° + θ) = - cot θ
csc (90° + θ) = sec θ
sec ( 90° + θ) = - csc θ
cot ( 90° + θ) = - tan θ
and
sin (180° + θ) = - sin θ
cos (180° + θ) = - cos θ
tan (180° + θ) = tan θ
csc (180° + θ) = -csc θ
sec (180° + θ) = - sec θ
cot (180° + θ) = cot θ
Using the above proved results we will prove all six trigonometrical ratios of (180° - θ).
sin (270° + θ) = sin [1800 + 90° + θ]
= sin [1800 + (90° + θ)]
= - sin (90° + θ), [since sin (180° + θ) = - sin θ]
Therefore, sin (270° + θ) = - cos θ, [since sin (90° + θ) = cos θ]

cos (270° + θ) = cos [1800 + 90° + θ]
= cos [I 800 + (90° + θ)]
= - cos (90° + θ), [since cos (180° + θ) = - cos θ]
Therefore, cos (270° + θ) = sin θ, [since cos (90° + θ) = - sin θ]

tan ( 270° + θ) = tan [1800 + 90° + θ]
= tan [180° + (90° + θ)]
= tan (90° + θ), [since tan (180° + θ) = tan θ]
Therefore, tan (270° + θ) = - cot θ, [since tan (90° + θ) = - cot θ]

csc (270° + θ) = 1sin(270°+Θ)1sin270Θ
= 1−cosΘ1cosΘ, [since sin (270° + θ) = - cos θ]
Therefore, csc (270° + θ) = - sec θ;

sec (270° + θ) =1cos(270°+Θ)1cos270Θ
= 1sinΘ1sinΘ, [since cos (270° + θ) = sin θ]
Therefore, sec (270° + θ) = csc θ
and
cot (270° + θ) = 1tan(270°+Θ)1tan270Θ
= 1−cotΘ1cotΘ, [since tan (270° + θ) = - cot θ]
Therefore, cot (270° + θ) = - tan θ.
Solved examples:
1. Find the value of csc 315°.
Solution:
csc 315° = sec (270 + 45)°
= - sec 45°; since we know, csc (270° + θ) = - sec θ
= - √2
2. Find the value of cos 330°.
Solution:
cos 330° = cos (270 + 60)°
= sin 60°; since we know, cos (270° + θ) = sin θ
= √3232

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Tips for each of the different sections of a Bank PO exam?

Ans. Aspirants should pay enough attention to each of the sections. For the Awareness section, they need to keep themselves updated about current happenings of the world and having knowledge of static General Knowledge would be enough from the examination point of view.

Coming to the Mathematics section, candidates must remember it is essentially a 10th level questions. Aspirants should learn to find shortcut methods and learn to identify the answers quickly from the given answer options. Bank PO aspirants should never get into the habit of solving a problem by writing the entire equation, as with this approach they wouldn’t be able to crack the exam.

In the English Section, vocabulary plays a very crucial role. As poor vocabulary can hinder the process of preparation. Candidates need to be at least average or above average in English to enhance their chances of succeeding in the exam. People who are poor in vocabulary are going to struggle in the PO test.

There are some questions in the Reasoning Section which doesn’t take much time to solve e.g. if a candidate solves 100-150 questions in ‘Series’ topic, he/she would practically cover all the types of questions that comes in Bank PO exam from ‘Series’ topic. What aspirants can do is schedule two days for each of the reasoning topics and start solving it. One of the plus point is, pattern of the exam doesn’t change much (at max it would be 10%). However, level of difficulty may vary up and down. Still there is not too much of surprise students can expect or get.

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